Free math worksheets

- Begin with the absolute-value series rather than the alternating-series test. - Identify the exponent in the resulting \(p\)-series. - Convergence of the absolute-value series gives the strongest classification directly.

1. The absolute-value series is \(\sum1/n^{3/2}\). 2. This is a convergent \(p\)-series because \(p=\frac32>1\). 3. Therefore the original series converges absolutely, which is the strongest classification.

The series converges absolutely.

- One definition is based entirely on the absolute-value series. - The other compares the outcomes of the signed and absolute-value series. - Conditional convergence requires the signed series itself to converge.

1. The series \(\sum a_n\) converges absolutely if \(\sum|a_n|\) converges. 2. It converges conditionally if \(\sum a_n\) converges but \(\sum|a_n|\) diverges.

Absolute convergence means \(\sum|a_n|\) converges. Conditional convergence means \(\sum a_n\) converges but \(\sum|a_n|\) diverges.

- Apply the alternating series test to the signed series. - Taking absolute values removes the alternating sign and produces a familiar benchmark series. - Conditional convergence requires convergence of the signed series but divergence of the absolute-value series.

1. For the signed series, the magnitudes \(1/n\) decrease to \(0\), so the series converges by the alternating series test. 2. The absolute-value series is the harmonic series \(\sum1/n\), which diverges. 3. Therefore the original series is conditionally convergent.

The signed series converges, but the absolute-value series diverges; therefore the series converges conditionally.

- First verify the alternating-series conditions for the signed series. - Then remove the signs and identify the exponent in the resulting \(p\)-series. - Absolute convergence requires the absolute-value series to converge.

1. For the signed series, the magnitudes \(1/n^2\) decrease to \(0\), so the alternating series test gives convergence. 2. The absolute-value series is \(\sum1/n^2\), a convergent \(p\)-series with \(p=2\). 3. Therefore the original series converges absolutely.

Both the signed series and the absolute-value series converge, so the series converges absolutely.

- Rewrite \(1/\sqrt n\) as a reciprocal power. - Use different tests for the signed series and the absolute-value series. - Compare the absolute-value exponent with the \(p\)-series boundary \(p=1\).

1. The signed series converges by the alternating series test because \(1/\sqrt n\) decreases to \(0\). 2. The absolute-value series is the divergent \(p\)-series \(\sum1/n^{1/2}\). 3. Therefore the original series is conditionally convergent.

The signed series converges, while the absolute-value series diverges; therefore the series converges conditionally.

- Check the signed series with the alternating series test. - Recognize the absolute-value series as a harmonic tail. - Replacing finitely many harmonic terms does not change divergence.

1. The magnitudes \(1/(n+1)\) decrease to \(0\), so the signed series converges by the alternating series test. 2. Replacing each term by its magnitude gives the shifted harmonic series \(\sum1/(n+1)\), which diverges. 3. Therefore the original series is conditionally convergent.

No. The signed series converges, but the absolute-value series diverges, so the original series converges conditionally.

- Combine \(3^n/2^n\) into \((3/2)^n\). - Compare consecutive magnitudes to determine whether the exponential growth is offset by \(n\). - Check the necessary term-limit condition before applying any convergence test.

1. The term magnitude is \(b_n=\frac{(3/2)^n}{n}\). 2. Its consecutive ratio is \(\frac{b_{n+1}}{b_n}=\frac32\frac{n}{n+1}\to\frac32>1\). 3. Thus the magnitudes eventually grow and do not approach \(0\). 4. Therefore the series diverges by the \(n\)th-term test for divergence, so no absolute-versus-conditional classification is needed.

The terms do not approach \(0\), so the series diverges.

- Evaluate \(\cos(\pi n)\) for integer values of \(n\). - Take absolute values after replacing the trigonometric factor. - Identify the resulting reciprocal-power series.

1. For integer \(n\), \(\cos(\pi n)=(-1)^n\). 2. Taking absolute values gives the \(p\)-series \(\sum1/n^2\), which converges. 3. Therefore the original series converges absolutely.

Since \(\cos(\pi n)=(-1)^n\) and \(\sum1/n^2\) converges, the series converges absolutely.

- Compare \(n^2+n\) with \(n^2\). - Remember that a larger positive denominator gives a smaller reciprocal. - Apply direct comparison to the absolute-value series.

1. The absolute terms satisfy \(0<\frac1{n^2+n}\le\frac1{n^2}\) because \(n^2+n\ge n^2\). 2. The benchmark \(\sum1/n^2\) converges. 3. Therefore the absolute-value series converges by direct comparison, so the original series converges absolutely.

The magnitudes are bounded above by \(1/n^2\), so the series converges absolutely.

- Recall the rearrangement theorem for absolutely convergent series. - Distinguish preservation of convergence from preservation of the sum. - State what happens for every ordering, not just one rearrangement.

1. Absolute convergence is preserved under rearrangement. 2. Every rearrangement converges to the same value \(S\).

Every rearrangement converges absolutely and has the same sum \(S\).

- Identify the hypothesis required by the same-sum rearrangement theorem. - Compare that hypothesis with the definition of conditional convergence. - The absence of a guarantee is enough; no particular rearrangement needs to be constructed.

1. The same-sum rearrangement guarantee applies to absolutely convergent series. 2. A conditionally convergent series is not absolutely convergent. 3. Therefore that guarantee does not apply, so a rearrangement is not guaranteed to have the same sum.

No. The same-sum rearrangement guarantee is not available for a conditionally convergent series.

- Use the constant-multiple rule for the signed series. - Compute \(|-4a_n|\) before considering absolute convergence. - Multiplication by a positive nonzero constant does not turn a divergent positive series into a convergent one.

1. Multiplying every term by the nonzero constant \(-4\) preserves convergence of the signed series. 2. The absolute-value series becomes \(\sum|-4a_n|=4\sum|a_n|\), which still diverges. 3. Therefore \(\sum(-4a_n)\) is conditionally convergent.

The series is conditionally convergent.

- Use the \(p\)-series boundary \(p=1\) to choose two exponents with different absolute behavior. - For the absolute example, choose an exponent greater than \(1\). - For the conditional example, choose a positive exponent whose absolute-value \(p\)-series diverges.

1. Choose \(p=2\). Then \(\sum_{n=1}^{\infty}(-1)^n/n^2\) converges absolutely because \(\sum_{n=1}^{\infty}1/n^2\) converges. 2. Choose \(p=1\). Then \(\sum_{n=1}^{\infty}(-1)^n/n\) converges by the alternating series test, while \(\sum_{n=1}^{\infty}1/n\) diverges, so it converges conditionally.

One valid pair is \(\sum_{n=1}^{\infty}(-1)^n/n^2\), which converges absolutely, and \(\sum_{n=1}^{\infty}(-1)^n/n\), which converges conditionally.

- How large can the oscillating numerator be in absolute value? - Compare the resulting magnitudes with a simpler positive sequence. - Use the magnitude behavior to state the strongest classification.

1. Since \(|\sin n|\le1\), \(\left|\frac{\sin n}{n^2}\right|\le\frac{1}{n^2}\). 2. The comparison series \(\sum1/n^2\) converges. 3. Therefore \(\sum|\sin n|/n^2\) converges by direct comparison. 4. Hence the original series converges absolutely.

The series is absolutely convergent.

- Start with a familiar convergent power series. - Choose its exponent so that taking a square root reaches the convergence boundary. - Verify the original and transformed series separately.

1. Choose \(a_n=1/n^2\). 2. The series \(\sum|a_n|=\sum1/n^2\) converges because it is a \(p\)-series with \(p=2\). 3. However, \(\sqrt{|a_n|}=1/n\). 4. The transformed series is the harmonic series, so it diverges.

One valid construction is \(a_n=\frac{1}{n^2}\). Then \(\sum a_n\) converges absolutely, but \(\sum\sqrt{|a_n|}=\sum1/n\) diverges.

- Begin with the magnitude of the product term. - Use the uniform bound that applies to every multiplier. - Compare with a constant multiple of the known series of absolute values.

1. For every \(n\), \(|c_n a_n|=|c_n||a_n|\le M|a_n|\). 2. The comparison series \(\sum M|a_n|\) converges because it is a constant multiple of the convergent series \(\sum|a_n|\). 3. Therefore \(\sum|c_n a_n|\) converges by direct comparison. 4. Hence \(\sum c_n a_n\) converges absolutely.

\(\sum c_n a_n\) must converge absolutely.

- Determine the sign of the expression inside the parentheses. - Write several consecutive magnitude terms and look for cancellation. - Use the behavior of the series of absolute values to classify the signed series.

1. The expression in parentheses is positive, so the series of absolute values is \(\sum_{n=1}^{\infty}(1/\sqrt n-1/\sqrt{n+1})\). 2. Its \(N\)th partial sum telescopes to \(1-1/\sqrt{N+1}\). 3. These partial sums approach \(1\), so the series of absolute values converges. 4. Therefore the original signed series converges absolutely.

The series is absolutely convergent.

- Bound the larger of two nonnegative numbers by a simple expression involving both. - Combine the two known series of absolute values. - Apply a comparison to the newly defined terms.

1. For every \(n\), \(0\le c_n\le|a_n|+|b_n|\). 2. The series \(\sum(|a_n|+|b_n|)\) converges because it is the sum of two convergent series of absolute values. 3. Therefore \(\sum c_n\) converges by direct comparison.

\(\sum c_n\) converges.

- Notice that no sign cancellation is needed for this series. - Compare the small-angle expression with its input. - Square the familiar trigonometric limit carefully.

1. The terms are nonnegative, so convergence is the same as absolute convergence. 2. Compare with \(1/n^2\): \(\frac{\sin^2(1/n)}{1/n^2}=\left(\frac{\sin(1/n)}{1/n}\right)^2\to1\). 3. Since \(\sum1/n^2\) converges, the given series converges by limit comparison. 4. Therefore it is absolutely convergent.

The series is absolutely convergent.

- Compare the magnitudes with the supplied geometric expression. - Separate the finite initial segment from the controlled tail. - Use the definition that distinguishes absolute from conditional convergence.

1. The geometric tail \(\sum_{n=N}^{\infty}Cr^n\) converges because \(0<r<1\). 2. Since \(|a_n|\le Cr^n\) for \(n\ge N\), the tail \(\sum_{n=N}^{\infty}|a_n|\) converges by direct comparison. 3. Adding the finite initial magnitude sum shows that \(\sum|a_n|\) converges. 4. Therefore \(\sum a_n\) converges absolutely, so conditional convergence is impossible.

The stated geometric bound forces absolute convergence, so conditional convergence is impossible.

- Begin with the absolute-value series. - Use the substitution \(u=\ln x\) in its improper integral. - Convergence after taking absolute values gives the final classification immediately.

1. The absolute-value series is \(\sum_{n=2}^{\infty}\frac1{n(\ln n)^2}\). 2. The integral test applies, and \(\int_2^\infty\frac{dx}{x(\ln x)^2}=\frac1{\ln2}<\infty\). 3. Therefore the absolute-value series converges, which also guarantees convergence of the signed series. 4. The original series is absolutely convergent.

The absolute-value series converges, so the signed series converges absolutely.

- Split \((n+1)/n^2\) into two reciprocal powers. - Use the component behavior to check the alternating-series conditions. - For absolute convergence, compare the sum of the components with \(1/n\).

1. The positive magnitude is \(\frac{n+1}{n^2}=\frac1n+\frac1{n^2}\). 2. Both components decrease to \(0\), so the magnitude decreases to \(0\) and the signed series converges by the alternating series test. 3. The absolute-value series diverges because it is at least the harmonic series termwise: \(\frac1n+\frac1{n^2}\ge\frac1n\). 4. Therefore the original series is conditionally convergent.

The signed series converges, but the absolute-value series diverges; therefore the series converges conditionally.

- Divide each numerator term by \(n^3\) separately. - Classify the two resulting reciprocal-power series. - A finite sum of convergent positive series also converges.

1. The absolute-value terms satisfy \(\frac{n+1}{n^3}=\frac1{n^2}+\frac1{n^3}\). 2. Both component \(p\)-series converge because their exponents exceed \(1\). 3. Therefore the absolute-value series converges, so the original series converges absolutely.

The absolute-value series is the sum of convergent \(p\)-series with exponents \(2\) and \(3\), so the series converges absolutely.

- Use a derivative to verify decrease of \(n/(n^2+1)\). - For absolute convergence, compare the magnitude with \(1/n\). - Compute the ratio of those two positive terms before classifying.

1. The magnitude \(b_n=\frac{n}{n^2+1}\) approaches \(0\). 2. For \(f(x)=\frac{x}{x^2+1}\), \(f'(x)=\frac{1-x^2}{(x^2+1)^2}\le0\) for \(x\ge1\), so the signed series converges by the alternating series test. 3. For absolute values, \(\lim_{n\to\infty}\frac{n/(n^2+1)}{1/n}=\lim_{n\to\infty}\frac{n^2}{n^2+1}=1\). 4. Thus the absolute-value series diverges by limit comparison with the harmonic series, so the original series is conditionally convergent.

The signed series converges, but the absolute-value series diverges by harmonic comparison; therefore the series converges conditionally.

- Use the leading powers to predict the reciprocal-square benchmark. - Divide the magnitude by \(1/n^2\) and simplify. - A positive finite ratio limit transfers the benchmark’s convergence.

1. The absolute-value term is \(b_n=\frac{n}{n^3+1}\). 2. Comparing with \(1/n^2\) gives \(\lim_{n\to\infty}\frac{b_n}{1/n^2}=\lim_{n\to\infty}\frac{n^3}{n^3+1}=1\). 3. Since \(\sum1/n^2\) converges, the absolute-value series converges by the limit comparison test. 4. Therefore the original series converges absolutely.

The absolute-value series has the same behavior as \(\sum1/n^2\), so the series converges absolutely.

- Verify that \(1/\sqrt{n^2+1}\) decreases to \(0\). - For absolute convergence, compare the magnitude with \(1/n\). - Simplify the comparison ratio by factoring \(n^2\) from the square root.

1. The magnitudes \(b_n=\frac1{\sqrt{n^2+1}}\) decrease to \(0\), so the signed series converges by the alternating series test. 2. For absolute values, \(\lim_{n\to\infty}\frac{1/\sqrt{n^2+1}}{1/n}=\lim_{n\to\infty}\frac{n}{\sqrt{n^2+1}}=1\). 3. Thus the absolute-value series diverges by limit comparison with the harmonic series. 4. Therefore the original series is conditionally convergent.

The signed series converges, but the absolute-value series diverges by harmonic limit comparison; therefore the series converges conditionally.

- Use the dominant power inside \(\sqrt{n^3+1}\) to choose a benchmark. - Compare with the \(p\)-series having exponent \(3/2\). - A convergent absolute-value series establishes absolute convergence directly.

1. For the absolute-value terms, compare \(b_n=\frac1{\sqrt{n^3+1}}\) with \(1/n^{3/2}\). 2. The ratio satisfies \(\lim_{n\to\infty}\frac{b_n}{1/n^{3/2}}=\lim_{n\to\infty}\frac{n^{3/2}}{\sqrt{n^3+1}}=1\). 3. Since \(\sum1/n^{3/2}\) converges, the absolute-value series converges by limit comparison. 4. Therefore taking absolute values does not destroy convergence; the original series converges absolutely.

Taking absolute values still gives a convergent series, so the original series converges absolutely.

- Combine the powers of \(2\) and \(3\) into one geometric factor. - Use the simple bound \(1/n\le1\). - Apply direct comparison to the absolute-value series.

1. The absolute-value terms are \(\frac{(2/3)^n}{n}\). 2. Since \(0<1/n\le1\), \(0<\frac{(2/3)^n}{n}\le(2/3)^n\). 3. The geometric benchmark \(\sum(2/3)^n\) converges, so the absolute-value series converges by direct comparison. 4. Therefore the original series converges absolutely.

The absolute-value series is bounded above by a convergent geometric series, so the original series converges absolutely.

- Recall the triangle inequality relating a signed sum to the sum of magnitudes. - Look for an example containing both a positive and a negative term. - Distinguish finiteness of \(\sum|a_n|\) from equality of two numerical sums.

1. Absolute convergence means that \(\sum|a_n|\) is finite; it does not assert equality with \(|\sum a_n|\). 2. The triangle inequality gives \(|\sum a_n|\le\sum|a_n|\), and cancellation can make the inequality strict. 3. For example, let \(a_1=1\), \(a_2=-\frac12\), and \(a_n=0\) for \(n\ge3\). Then \(|\sum a_n|=\frac12\), while \(\sum|a_n|=\frac32\).

Absolute convergence guarantees that \(\sum|a_n|\) is finite, not that it equals \(|\sum a_n|\); cancellation can make the latter smaller.

- Separate the effect on the signed series from the effect on the absolute-value series. - Changing finitely many terms cannot alter convergence of the signed series. - A sign change leaves each term’s magnitude unchanged.

1. Changing the signs of finitely many terms changes the signed sum by only a finite amount, so convergence is preserved. 2. The absolute values of those terms do not change at all, so the absolute-value series remains divergent. 3. Therefore the new series remains conditionally convergent.

The new series is conditionally convergent.

- Use the difference-of-cubes identity to study the positive magnitudes. - Check the signed series using the monotonicity of the rewritten denominator. - Analyze the series of absolute values through its finite partial sums.

1. The magnitude can be rewritten as \(b_n=1/[(n+1)^{2/3}+(n(n+1))^{1/3}+n^{2/3}]\). 2. Its denominator increases, so \(b_n\) decreases, and \(b_n\to0\). The signed series converges by the alternating series test. 3. The partial sums of the series of absolute values telescope: \(\sum_{n=1}^{N}(\sqrt[3]{n+1}-\sqrt[3]{n})=\sqrt[3]{N+1}-1\). 4. These partial sums grow without bound, so the series of absolute values diverges. 5. Therefore the original series converges conditionally.

The series is conditionally convergent.

- Use what convergence of the series of absolute values implies about the individual terms. - Compare a small number with its square. - A finite initial segment does not affect convergence.

1. Absolute convergence implies \(a_n\to0\). 2. Therefore there is an index \(N\) such that \(|a_n|\le1\) for every \(n\ge N\). 3. For those indices, \(|a_n^2|=|a_n|^2\le|a_n|\). 4. The tail \(\sum_{n=N}^{\infty}|a_n|\) converges, so \(\sum_{n=N}^{\infty}|a_n|^2\) converges by comparison. 5. Adding the finite initial portion shows that \(\sum a_n^2\) converges absolutely.

\(\sum a_n^2\) must converge absolutely.

- Find a convenient upper estimate for the running reciprocal sum. - Can the slowly growing part be bounded by a small positive power for large indices? - Use the resulting bound on the full magnitudes.

1. The harmonic-number bound \(H_n\le1+\ln n\) holds for \(n\ge1\). 2. Since \(\ln n\le\sqrt n\) for \(n\ge1\), \(H_n\le1+\sqrt n\le2\sqrt n\). 3. Therefore \(\left|(-1)^nH_n/n^2\right|\le2/n^{3/2}\). 4. The comparison \(p\)-series converges, so the given series converges absolutely.

The series is absolutely convergent.

- Rewrite the radical difference without subtraction. - Bound the resulting denominator using two copies of the smaller radical. - Compare the cubed magnitude with a power series.

1. Rationalizing gives \(\sqrt{n+1}-\sqrt n=1/(\sqrt{n+1}+\sqrt n)\). 2. Therefore \(\left|(-1)^n(\sqrt{n+1}-\sqrt n)^3\right|=1/(\sqrt{n+1}+\sqrt n)^3\). 3. Since \(\sqrt{n+1}+\sqrt n\ge2\sqrt n\), the magnitude is at most \(1/(8n^{3/2})\). 4. The comparison \(p\)-series converges, so the given series converges absolutely.

The series is absolutely convergent.

- View the prime-indexed terms as a selection from a nonnegative convergent series. - Compare each finite selected sum with a finite initial sum of the full series of absolute values. - Use bounded monotone partial sums.

1. Absolute convergence gives \(\sum_{n=1}^{\infty}|a_n|<\infty\). 2. Every term \(|a_{p_k}|\) is one of the nonnegative terms in the full series of absolute values. 3. For each \(K\), \(\sum_{k=1}^{K}|a_{p_k}|\le\sum_{n=1}^{p_K}|a_n|\le\sum_{n=1}^{\infty}|a_n|\). 4. The partial sums of the prime-indexed series of absolute values are increasing and bounded above, so they converge. 5. Hence \(\sum a_{p_k}\) converges absolutely.

The prime-indexed subseries must converge absolutely.

- Evaluate each integral or combine adjacent integration intervals. - Check whether the positive magnitudes decrease to zero. - Compare the signed and magnitude partial sums separately.

1. The magnitude is \(b_n=\int_n^{n+1}1/x\,dx=\ln((n+1)/n)=\ln(1+1/n)\). 2. The sequence \(b_n\) decreases to \(0\), so the signed series converges by the alternating series test. 3. The \(N\)th partial sum of the series of absolute values is \(\sum_{n=1}^{N}\int_n^{n+1}1/x\,dx=\int_1^{N+1}1/x\,dx=\ln(N+1)\). 4. These partial sums grow without bound, so the series of absolute values diverges. 5. Therefore the original series converges conditionally.

The series is conditionally convergent.

- Use both consequences contained in the phrase “conditionally convergent.” - Once the terms are small, bound the new denominator by a constant. - Compare the transformed terms from below with the original magnitudes.

1. Conditional convergence implies \(a_n\to0\) and \(\sum|a_n|\) diverges. 2. There is an index \(N\) such that \(|a_n|\le1\) for every \(n\ge N\). 3. For \(n\ge N\), \(\frac{|a_n|}{1+|a_n|}\ge\frac{|a_n|}{2}\). 4. Since \(\sum_{n=N}^{\infty}|a_n|/2\) diverges, the transformed positive series diverges by comparison.

\(\sum |a_n|/(1+|a_n|)\) must diverge.

- Evaluate the trigonometric factor separately at odd and even indices. - Remove the zero terms and rewrite what remains as a series indexed by half the original index. - Compare the signed behavior with the behavior of the magnitudes.

1. For odd \(n\), \(\cos(n\pi/2)=0\). 2. For \(n=2k\), \(\cos(n\pi/2)=\cos(k\pi)=(-1)^k\). Thus the series reduces to \(\sum_{k=1}^{\infty}(-1)^k/\sqrt{2k}\). 3. The positive magnitudes decrease to \(0\), so the reduced signed series converges. 4. The series of absolute values is \(\sum_{k=1}^{\infty}1/\sqrt{2k}\), which diverges. 5. Therefore the original series converges conditionally.

The series is conditionally convergent.

- Represent the positive and negative contributions with two nonnegative unknowns. - Write one equation for the signed total and another for the absolute total. - Combine the equations to isolate each contribution.

1. Let \(P\) be the sum of the positive terms and \(N\) the sum of the magnitudes of the negative terms. 2. The signed sum gives \(P-N=3\). 3. The absolute sum gives \(P+N=11\). 4. Adding the equations gives \(2P=14\), so \(P=7\). 5. Substituting into either equation gives \(N=4\).

The positive terms sum to \(7\), and the magnitudes of the negative terms sum to \(4\).

- Separate the partial sums according to whether the number of included terms is even or odd. - Relate each parity case to the two original partial-sum sequences. - For absolute convergence, inspect a subseries consisting of only one set of positions.

1. The even partial sums satisfy \(\sum_{k=1}^{2N}c_k=\sum_{n=1}^{N}a_n+\sum_{n=1}^{N}b_n\to A+B\). 2. The odd partial sums satisfy \(\sum_{k=1}^{2N-1}c_k=\sum_{n=1}^{N}a_n+\sum_{n=1}^{N-1}b_n\to A+B\). 3. Because both parity subsequences of partial sums have the same limit, \(\sum c_n\) converges to \(A+B\). 4. The series of absolute values contains the odd-indexed subseries \(\sum |c_{2n-1}|=\sum|a_n|\), which diverges. 5. Therefore \(\sum c_n\) is not absolutely convergent and is conditionally convergent.

The interleaved series converges conditionally, and its sum is \(A+B\).

- Compare the highest powers in both the numerator and denominator. - Verify that the positive magnitudes eventually decrease. - Test the series of absolute values against the correct power benchmark.

1. The numerator also grows with \(n\), so the magnitude behaves like \(n/n^2=1/n\), not like \(1/n^2\). 2. For \(f(x)=(x+2)/(x^2+3x+5)\), \(f'(x)=-(x^2+4x+1)/(x^2+3x+5)^2<0\) for \(x>0\). 3. The magnitudes decrease to \(0\), so the signed series converges by the alternating series test. 4. Also \(\lim_{n\to\infty}\frac{(n+2)/(n^2+3n+5)}{1/n}=1\). 5. The series of absolute values diverges by limit comparison with the harmonic series, so the original series converges conditionally.

Yui ignored the growth of the numerator. The series is conditionally convergent, not absolutely convergent.

- Assume one sign stops occurring after some index. - Compare the signed terms and their magnitudes on that tail. - Recall that changing or adding finitely many terms does not affect convergence classification.

1. Suppose instead that only finitely many negative terms occur. Then all terms are nonnegative after some index \(N\). 2. On that tail, \(|a_n|=a_n\), so convergence of \(\sum a_n\) implies convergence of \(\sum|a_n|\) after index \(N\). 3. Adding the finitely many initial magnitudes would make the full series absolutely convergent, contradicting conditional convergence. 4. The same argument rules out having only finitely many positive terms. 5. Therefore both signs must occur infinitely often.

Every conditionally convergent series has infinitely many positive terms and infinitely many negative terms.

- Compare sufficiently long partial sums of the signed series. - Repeat the comparison for the series of absolute values. - A finite reordering changes neither accumulated totals nor convergence behavior.

1. For every partial sum beyond index \(30\), the new series contains exactly the same first \(n\) terms as the original series, only with a finite initial reordering. 2. Thus the two partial sums are equal for all \(n\ge30\), so the new series has the same sum. 3. The series of absolute values is also unchanged term for term except for the order of its first \(30\) terms. 4. Since the original series of absolute values diverges, the new one diverges. Therefore the new series remains conditionally convergent.

The new series is conditionally convergent and has the same sum as the original series.

- For each series, first ask whether testing absolute values gives a quick conclusion. - If absolute convergence fails or is unclear, inspect the sign pattern and magnitude behavior. - Before using any more elaborate test, check whether the terms of a candidate series even approach \(0\).

1. For the first series, \(1\le2+\cos n\le3\), so the absolute values are at most \(3/n^2\). Thus the first series converges absolutely. 2. For the second series, the magnitudes satisfy \((n+1)/n^2=1/n+1/n^2\), which are positive, decreasing, and approach \(0\). Hence the signed series converges by the alternating series test. Its absolute-value series contains the divergent harmonic component \(\sum1/n\), so it is conditionally convergent. 3. For the third series, \((n^2+1)/(n^2+3)\to1\). Therefore the signed terms do not approach \(0\), so the series diverges.

a) Absolutely convergent. b) Conditionally convergent. c) Divergent.

- Use the derivative of \(\ln x/x\) to verify decrease for the signed-series test. - For absolute convergence, integrate \(\ln x/x\) using the substitution \(u=\ln x\). - Conditional convergence requires opposite outcomes for the signed and absolute-value series.

1. The magnitudes \(b_n=\frac{\ln n}{n}\) approach \(0\) and decrease for \(n\ge3\), so the signed series converges by the alternating series test. 2. For the absolute-value series, \(\int_3^R\frac{\ln x}{x}\,dx=\frac12\big((\ln R)^2-(\ln3)^2\big)\to\infty\). 3. Thus the absolute-value series diverges by the integral test. 4. Therefore the original series is conditionally convergent.

The signed series converges, while the absolute-value series diverges; therefore the series converges conditionally.

- Remove the alternating sign and analyze \(\ln n/n^2\). - For the integral test, use integration by parts on \(\int \ln x/x^2\,dx\). - Once the absolute-value series converges, no separate alternating-series test is needed.

1. The absolute-value series is \(\sum_{n=2}^{\infty}\frac{\ln n}{n^2}\). 2. The associated function is positive, continuous, and eventually decreasing. 3. Integration by parts gives \(\int_2^\infty\frac{\ln x}{x^2}\,dx=\frac{\ln2+1}{2}<\infty\). 4. Therefore the absolute-value series converges, so the original series converges absolutely.

The absolute-value series converges, so the original series converges absolutely.

- Verify that \(1/(n\ln n)\) decreases to \(0\) for the alternating-series test. - Test absolute convergence with the substitution \(u=\ln x\). - Use the two outcomes to choose between absolute and conditional convergence.

1. The positive magnitudes \(b_n=\frac1{n\ln n}\) decrease to \(0\), so the signed series converges by the alternating series test. 2. For the absolute-value series, \(\int_2^R\frac{dx}{x\ln x}=\ln(\ln R)-\ln(\ln2)\to\infty\). 3. Thus the absolute-value series diverges by the integral test. 4. Therefore the original series is conditionally convergent.

The signed series converges, but \(\sum|a_n|\) diverges; therefore the series converges conditionally.

- Evaluate \(\sin(\pi n/2)\) for one complete four-term cycle. - Remove the zero terms and identify the alternating series that remains. - After taking absolute values, compare the odd reciprocals with the harmonic series.

1. The sine values cycle through \(1,0,-1,0\), so the nonzero terms form \(1-\frac13+\frac15-\frac17+\cdots\). 2. This alternating odd-reciprocal series converges because its magnitudes decrease to \(0\). 3. After taking absolute values, the nonzero terms form \(1+\frac13+\frac15+\frac17+\cdots\), which diverges as the odd-term subseries of the harmonic series. 4. Therefore the original series is conditionally convergent.

The signed series converges, but its absolute-value series diverges; therefore the series converges conditionally.

- Use the \(p\)-series boundary \(p=1\) for absolute convergence. - Separately determine when \(1/n^p\) decreases to \(0\). - Use the term-limit condition when \(p\) is nonpositive.

1. If \(p>1\), the absolute-value \(p\)-series converges, so the original series converges absolutely. 2. If \(0<p\le1\), the magnitudes decrease to \(0\), so the signed series converges by the alternating series test, while the absolute-value \(p\)-series diverges. Thus convergence is conditional. 3. If \(p\le0\), the terms do not approach \(0\), so the series diverges.

Absolutely convergent for \(p>1\). Conditionally convergent for \(0<p\le1\). Divergent for \(p\le0\).

- Replace the exponent \(2q\) by a single temporary \(p\)-series exponent. - Translate the absolute and conditional \(p\)-ranges back into inequalities for \(q\). - Check the zero-term condition when the effective exponent is nonpositive.

1. The effective reciprocal-power exponent is \(p=2q\). 2. Absolute convergence requires \(2q>1\), which gives \(q>\frac12\). 3. Conditional convergence requires \(0<2q\le1\), which gives \(0<q\le\frac12\). 4. If \(q\le0\), the terms do not approach \(0\), so the series diverges.

Absolutely convergent for \(q>\frac12\). Conditionally convergent for \(0<q\le\frac12\). Divergent for \(q\le0\).

- Separate the cases below, at, and above the geometric boundary \(c=1\). - For \(c<1\), compare the magnitudes with \(c^n\). - For \(c>1\), check the term limit before applying any series test.

1. If \(0\le c<1\), then \(0\le c^n/n\le c^n\), and the geometric benchmark converges. Thus the series converges absolutely. 2. If \(c=1\), the series is the alternating harmonic series, so it converges conditionally. 3. If \(c>1\), the term magnitudes \(c^n/n\) do not approach \(0\), so the series diverges.

Absolutely convergent for \(0\le c<1\). Conditionally convergent for \(c=1\). Divergent for \(c>1\).

- First determine when \(1/(n^p+1)\) decreases to \(0\). - For absolute convergence, use limit comparison with \(1/n^p\). - Treat nonpositive \(p\) with the necessary term-limit condition.

1. If \(p>0\), the magnitudes \(b_n=\frac1{n^p+1}\) decrease to \(0\), so the signed series converges by the alternating series test. 2. For absolute convergence, \(\lim_{n\to\infty}\frac{1/(n^p+1)}{1/n^p}=1\). Thus the absolute-value series has the same behavior as \(\sum1/n^p\). 3. Therefore convergence is absolute for \(p>1\) and conditional for \(0<p\le1\). 4. If \(p\le0\), the terms do not approach \(0\), so the series diverges.

Absolutely convergent for \(p>1\). Conditionally convergent for \(0<p\le1\). Divergent for \(p\le0\).

- Use \(c>-1\) to ensure every shifted base \(n+c\) is positive. - Compare \((n+c)^{-p}\) with \(n^{-p}\) for large \(n\). - Apply the alternating-series and term-limit conditions in the remaining exponent ranges.

1. Since \(c>-1\), \(n+c>0\) for all \(n\ge1\), and the fixed shift does not change the tail’s reciprocal-power behavior. 2. If \(p>1\), the absolute-value series converges by limit comparison with \(\sum1/n^p\). 3. If \(0<p\le1\), the magnitudes decrease to \(0\), so the signed series converges, while the absolute-value series diverges. Thus convergence is conditional. 4. If \(p\le0\), the terms fail to approach \(0\), so the series diverges.

Absolutely convergent for \(p>1\). Conditionally convergent for \(0<p\le1\). Divergent for \(p\le0\).

- First use the sum rule to establish ordinary convergence. - To rule out absolute convergence, assume it and subtract the absolutely convergent \(b_n\) series. - Compare the resulting conclusion with the hypothesis that \(\sum a_n\) is only conditionally convergent.

1. Both \(\sum a_n\) and \(\sum b_n\) converge, so \(\sum(a_n+b_n)\) converges. 2. Assume for contradiction that \(\sum(a_n+b_n)\) converges absolutely. 3. Since \(\sum b_n\) converges absolutely, the difference \(\sum a_n=\sum(a_n+b_n)-\sum b_n\) would then be absolutely convergent. 4. This contradicts the conditional convergence of \(\sum a_n\). 5. Therefore \(\sum(a_n+b_n)\) converges but not absolutely, so it converges conditionally.

The series \(\sum(a_n+b_n)\) converges conditionally.

- Separate the given series into two familiar series before testing absolute convergence. - For the magnitudes, compare the smaller correction with the dominant reciprocal term. - A lower bound by a constant multiple of a familiar divergent series is enough.

1. The signed series is the sum of the alternating harmonic series and the convergent \(p\)-series \(\sum1/n^2\), so it converges. 2. For every \(n\ge2\), the two summands have opposite signs when \(n\) is even and the same sign when \(n\) is odd. 3. In either case, \(\left|\frac{(-1)^{n+1}}{n}+\frac{1}{n^2}\right|\ge\frac{1}{n}-\frac{1}{n^2}\ge\frac{1}{2n}\). 4. The series of absolute values diverges by comparison with the harmonic series. 5. Therefore the original series converges conditionally.

The series is conditionally convergent.

- Compare the new even partial sums with the original partial sums. - Express a new odd partial sum as a nearby original partial sum plus or minus one term. - Check absolute convergence separately by examining the reordered magnitudes.

1. The even partial sums of the new series satisfy \(\sum_{k=1}^{2N}c_k=\sum_{k=1}^{2N}a_k\to A\). 2. The odd partial sums satisfy \(\sum_{k=1}^{2N-1}c_k=\sum_{k=1}^{2N}a_k-a_{2N-1}\). 3. Since every convergent series has \(a_n\to0\), the odd partial sums also approach \(A\). 4. Thus the new series converges to \(A\). 5. Swapping adjacent terms does not change the absolute values as a collection. Its even absolute partial sums equal \(\sum_{k=1}^{2N}|a_k|\), which diverges. 6. Therefore the new series is conditionally convergent.

The adjacent-pair swap preserves the sum \(A\) and preserves conditional convergence.

- Begin with what linearity guarantees for any two convergent series. - To obtain conditional convergence, avoid cancellation of the dominant magnitudes. - To obtain absolute convergence, arrange cancellation of the conditionally convergent parts while leaving a smaller remainder.

1. The sum series cannot diverge because the sum of two convergent series is convergent. 2. Conditional convergence is possible. Take \(a_n=(-1)^{n+1}/n\) and \(b_n=(-1)^{n+1}/\sqrt n\). Both series converge conditionally, and \(|a_n+b_n|=1/n+1/\sqrt n\), so their sum is not absolutely convergent. 3. Absolute convergence is also possible. Take \(a_n=(-1)^{n+1}/n\) and \(b_n=(-1)^n(1/n+1/n^{3/2})\). Both series converge conditionally, but \(a_n+b_n=(-1)^n/n^{3/2}\), whose series converges absolutely. 4. Thus absolute and conditional convergence are possible, while divergence is impossible.

Absolutely convergent: possible. Conditionally convergent: possible. Divergent: impossible.

- Compare a grouped partial sum with an even partial sum of the original series. - Use the ordering of consecutive magnitudes to determine the sign of each grouped term. - Distinguish grouping terms before taking absolute values from taking absolute values first.

1. Since \(b_{2n-1}\ge b_{2n}\), each \(c_n\ge0\). 2. The \(N\)th partial sum of \(\sum c_n\) is \(\sum_{n=1}^{N}(b_{2n-1}-b_{2n})\), which is the \(2N\)th partial sum of the alternating series. 3. The alternating series converges, so these partial sums approach its sum. Therefore \(\sum c_n\) converges. 4. Because every \(c_n\ge0\), convergence of \(\sum c_n\) is absolute convergence. 5. The original series of absolute values is \(\sum b_n\), which diverges by assumption. 6. Grouping two original terms into one new term changes the term sequence whose absolute values are being summed, so absolute convergence of the grouped series does not imply absolute convergence of the original series.

The grouped series \(\sum c_n\) converges absolutely to the same sum as the alternating series. The original series remains conditionally convergent because \(\sum b_n\) diverges.

- What term behavior follows from absolute convergence? - Once the inputs are sufficiently small, how can the logarithmic change be bounded by the input size? - Use that tail bound to compare the magnitude totals.

1. Absolute convergence implies \(a_n\to0\). 2. Choose \(N\) so that \(|a_n|\le1/2\) for every \(n\ge N\). 3. By the mean value theorem applied to \(\ln x\) between \(1\) and \(1+a_n\), \(|\ln(1+a_n)|=|a_n|/\xi_n\) for some \(\xi_n\ge1/2\). 4. Hence \(|\ln(1+a_n)|\le2|a_n|\) for \(n\ge N\). 5. The series \(\sum|\ln(1+a_n)|\) converges by comparison with the convergent tail \(\sum2|a_n|\). Finite initial terms do not affect convergence.

\(\sum\ln(1+a_n)\) converges absolutely.

- Which denominator term dominates in each parameter range? - Check whether the full denominator eventually increases even when one component decreases. - Classify the signed series and the series of absolute values separately.

1. Let \(f(x)=x^p+\sqrt x\). If \(p\ge0\), then \(f\) is increasing. If \(p<0\), then \(f'(x)=x^{p-1}\left(p+\frac12x^{1/2-p}\right)>0\) for all sufficiently large \(x\). Thus the positive magnitudes are eventually decreasing. 2. Since \(n^p+\sqrt n\to\infty\) for every real \(p\), the magnitudes approach \(0\). Therefore the signed series converges for every real \(p\) by the alternating series test. 3. If \(p>1/2\), compare the absolute terms with \(1/n^p\): \(\frac{1/(n^p+\sqrt n)}{1/n^p}=\frac{1}{1+n^{1/2-p}}\to1\). 4. If \(p=1/2\), the absolute terms equal \(1/(2\sqrt n)\). If \(p<1/2\), compare with \(1/\sqrt n\): \(\frac{1/(n^p+\sqrt n)}{1/\sqrt n}=\frac{1}{1+n^{p-1/2}}\to1\). 5. The series of absolute values therefore converges exactly when \(p>1\). Hence the original series is absolutely convergent for \(p>1\) and conditionally convergent for \(p\le1\).

a) \(A(p)\) converges for every real \(p\). b) Its series of absolute values converges exactly when \(p>1\). Thus \(A(p)\) is absolutely convergent for \(p>1\) and conditionally convergent for \(p\le1\).

- Try matching the signs so the product terms become positive. - Choose exponents that are individually at or below the absolute-convergence boundary. - Make the sum of the two exponents fall on different sides of that boundary in two examples.

1. For divergence, choose \(a_n=b_n=(-1)^n/\sqrt n\). Each original series converges conditionally, but \(a_n b_n=1/n\), so the product series diverges. 2. For absolute convergence, choose \(a_n=b_n=(-1)^n/n^{2/3}\). Each original series converges conditionally, but \(|a_n b_n|=1/n^{4/3}\). 3. Since \(\sum1/n^{4/3}\) converges, this second product series converges absolutely. 4. Therefore conditional convergence of both factor series does not determine one fixed behavior for the product series.

Both outcomes are possible: the product series can diverge, or it can converge absolutely.

- Determine the sign of the difference before taking magnitudes. - After the leading small-input behavior cancels, what order of size remains? - Compare that remaining order with a familiar convergent benchmark.

1. Since \(\ln(1+x)<x\) for \(x>0\), each term is negative and its magnitude is \(1/n-\ln(1+1/n)\). 2. Set \(x=1/n\). Then \(\lim_{x\to0^+}\frac{x-\ln(1+x)}{x^2}=1/2\), obtained by differentiating numerator and denominator once. 3. Therefore \(\lim_{n\to\infty}\frac{1/n-\ln(1+1/n)}{1/n^2}=1/2\). 4. The series of absolute values converges by limit comparison with \(\sum1/n^2\). 5. Hence the original series converges absolutely.

The series converges absolutely.

- Express each term and each magnitude using its positive and negative parts. - Assume one of the two nonnegative series converges and use the convergence of the signed series. - Compare the conclusion with the definition of conditional convergence.

1. For every \(n\), \(a_n=a_n^+-a_n^-\) and \(|a_n|=a_n^++a_n^-\). 2. Suppose \(\sum a_n^+\) converged. Since \(\sum a_n\) also converges, the identity \(a_n^-=a_n^+-a_n\) would imply that \(\sum a_n^-\) converges. 3. Then \(\sum|a_n|=\sum(a_n^++a_n^-)\) would converge, contradicting conditional convergence. 4. Thus \(\sum a_n^+\) diverges. The same argument with the roles reversed shows that \(\sum a_n^-\) diverges.

Both the positive-part series \(\sum a_n^+\) and the negative-part series \(\sum a_n^-\) diverge.

- Which powers dominate the numerator and denominator at large indices? - Check whether the magnitudes eventually move downward and approach zero. - Then compare the magnitude size with the boundary reciprocal behavior.

1. The magnitudes approach \(0\) because \(\frac{n+a}{n^2+b}=\frac{1/n+a/n^2}{1+b/n^2}\to0\). 2. For \(f(x)=(x+a)/(x^2+b)\), \(f'(x)=\frac{-x^2-2ax+b}{(x^2+b)^2}\), which is negative for all sufficiently large \(x\). 3. Thus the signed series converges by the alternating series test after ignoring finitely many initial terms. 4. For absolute convergence, \(\lim_{n\to\infty}\frac{(n+a)/(n^2+b)}{1/n}=1\). 5. The series of absolute values diverges by limit comparison with the harmonic series, so the original series converges conditionally for every allowed \(a\) and \(b\).

The series is conditionally convergent for every \(a\ge0\) and \(b\ge0\).

- Compare even and odd partial sums of the new series separately. - A completed pair should reproduce one original term. - Compare the total magnitude of each pair with the original magnitude.

1. Let \(S_N=\sum_{n=1}^{N}a_n\) and \(T_M=\sum_{m=1}^{M}b_m\). 2. Each completed pair contributes \(a_n\), so \(T_{2N}=S_N\to S\). 3. Also \(T_{2N-1}=S_{N-1}+a_N/2\to S\) because \(S_{N-1}\to S\) and \(a_N\to0\). 4. Thus the full sequence of partial sums \(T_M\) approaches \(S\). 5. For absolute values, each pair contributes \(|a_n|\), so \(\sum|b_m|=\sum|a_n|\), which diverges. 6. Therefore \(\sum b_m\) converges conditionally to \(S\).

\(\sum b_m\) converges conditionally, and its sum is \(S\).

- State exactly what the ratio test says when its limiting quotient equals the boundary value. - After an inconclusive test, analyze the signed series and the absolute-value series separately. - For the signed series, inspect the magnitude sequence; for the absolute-value series, compare its dominant size with a standard reciprocal power.

1. A ratio-test limit of \(L=1\) is inconclusive. It does not prove that the absolute-value series diverges. 2. Let \(b_n=1/\sqrt{n+\sqrt n}\). The denominator increases with \(n\), so \((b_n)\) is decreasing, and \(b_n\to0\). 3. Therefore the signed series converges by the alternating series test. 4. For absolute convergence, compare with \(1/\sqrt n\): \(\lim_{n\to\infty}\frac{1/\sqrt{n+\sqrt n}}{1/\sqrt n}=\lim_{n\to\infty}\sqrt{\frac{n}{n+\sqrt n}}=1\). Since \(\sum 1/\sqrt n\) diverges, the absolute-value series diverges by limit comparison. 5. Hence the original series is conditionally convergent.

The error is treating the ratio-test result \(L=1\) as a divergence result. The ratio test is inconclusive at \(L=1\). The original series converges by the alternating series test, while its absolute-value series diverges by limit comparison with \(\sum 1/\sqrt n\). Therefore the series is conditionally convergent.

- First add all three terms in one complete block. - Then check whether stopping partway through a late block changes a partial sum significantly. - Analyze the magnitude contributed by an entire block separately.

1. The sum of block \(k\) is \(1/k+1/k-2/k+1/k^2=1/k^2\). 2. Partial sums at the ends of blocks therefore converge because \(\sum1/k^2\) converges. 3. Within block \(k\), the two incomplete-block changes are \(1/k\) and \(2/k\), both of which approach \(0\). Hence all partial sums approach the same limit, so the series converges. 4. The sum of the magnitudes in block \(k\) is \(1/k+1/k+(2/k-1/k^2)=4/k-1/k^2\). 5. These block magnitudes have a divergent harmonic component, so \(\sum|a_n|\) diverges. 6. Therefore the series converges conditionally.

The series is conditionally convergent.

- Compare each modified term with the corresponding original term. - The changed indices form a geometric pattern. - Check whether sign changes alter the magnitudes.

1. The original series \(\sum a_n\) converges conditionally. 2. Since \(1=2^0\), changing the sign of \(a_1=1\) adds \(-2\) to that term. 3. At \(n=2^k\) with \(k\ge1\), \(a_{2^k}=-1/2^k\), so changing its sign adds \(2/2^k\) to that term. 4. Therefore the correction series consists of the term \(-2\) and the geometric series \(\sum_{k=1}^{\infty}2/2^k\), so it converges absolutely. 5. The new series is the sum of a convergent series and an absolutely convergent correction, so \(\sum b_n\) converges. 6. Since \(|b_n|=|a_n|=1/n\) for every \(n\), the series of absolute values is harmonic and diverges. 7. Thus \(\sum b_n\) converges conditionally.

The modified series converges conditionally.

- Place the terms of a familiar conditionally convergent series at increasingly sparse indices. - Use zeros to control the lower limit. - Make the chosen indices grow much faster than the reciprocal magnitudes shrink.

1. Define \(a_{2^k}=(-1)^{k+1}/k\) for \(k\ge1\), and define \(a_n=0\) when \(n\) is not a power of \(2\). 2. Reading only the nonzero terms gives the alternating harmonic series \(\sum_{k=1}^{\infty}(-1)^{k+1}/k\), so \(\sum a_n\) converges. 3. The series of absolute values has nonzero terms \(1/k\), so \(\sum|a_n|\) diverges. Thus the series is conditionally convergent. 4. At every non-power of \(2\), \(n|a_n|=0\), so the lower limit is \(0\). 5. At \(n=2^k\), \(n|a_n|=2^k/k\to\infty\), so the upper limit is infinite.

One valid construction is \(a_{2^k}=(-1)^{k+1}/k\) and \(a_n=0\) otherwise.