Free math worksheets

- Match the exponent in the denominator with the standard form \(\frac1{n^p}\). - Recall the convergence cutoff for a \(p\)-series. - Compare the identified exponent with that cutoff.

1. The term already has the form \(\frac1{n^p}\) with \(p=3\). 2. A \(p\)-series converges when \(p>1\). 3. Since \(3>1\), the series converges.

\(p=3\), so the series converges.

- Identify the exponent \(p\) in the standard reciprocal-power form. - Recall which side of \(p=1\) gives convergence. - Compare \(\frac34\) with the boundary value before classifying the series.

1. Rewrite \(\frac1{n^{3/4}}=n^{-3/4}\). In \(p\)-series form, the exponent is \(p=\frac34\). 2. A \(p\)-series diverges when \(p\le1\). 3. Since \(\frac34\le1\), the series diverges.

\(p=\frac34\), so the series diverges.

- Read the exponent directly from the denominator’s power of \(n\). - Recall the strict inequality required for convergence of a \(p\)-series. - Compare the small positive exponent with the boundary value \(1\).

1. The summand is already in the form \(\frac1{n^p}\) with \(p=0.01\). 2. A \(p\)-series converges only when \(p>1\). 3. Since \(0.01\le1\), the series diverges.

\(p=0.01\), so the series diverges.

- Match \(\frac1n\) with the standard form \(\frac1{n^p}\). - Recall the special name for the boundary case \(p=1\). - Apply the strict \(p>1\) convergence condition.

1. The summand has the form \(\frac1{n^p}\) with \(p=1\). 2. The case \(p=1\) is the harmonic series. 3. Since a \(p\)-series converges only for \(p>1\), the harmonic series diverges.

It is the harmonic series with \(p=1\), and it diverges.

- Factor the constant \(5\) outside the series. - Identify the exponent in the remaining reciprocal-power series. - A nonzero constant multiple preserves convergence or divergence.

1. \(\sum_{n=1}^{\infty}\frac5n=5\sum_{n=1}^{\infty}\frac1n\), whose underlying \(p\)-series has \(p=1\). 2. The harmonic series diverges. 3. Multiplication by the nonzero constant \(5\) does not change that convergence behavior.

The underlying series has \(p=1\); the constant factor \(5\) does not change divergence.

- Read the exact fractional exponent from the denominator. - Recall that the convergence inequality for a \(p\)-series is strict. - Compare \(\frac{11}{10}\) with \(1\), not with a rounded decimal estimate.

1. The summand has the form \(\frac1{n^p}\) with \(p=\frac{11}{10}\). 2. A \(p\)-series converges when \(p>1\). 3. Since \(\frac{11}{10}>1\), the series converges.

\(p=\frac{11}{10}\), so the series converges.

- State the exact, strict inequality in the \(p\)-series convergence theorem. - Compare \(p=1\) with that inequality. - Recall the name and behavior of the boundary case.

1. Positivity of \(p\) is not the convergence criterion for a \(p\)-series. 2. A \(p\)-series converges only when \(p>1\). 3. At \(p=1\), the series is harmonic and diverges.

The student used the wrong cutoff. The harmonic case \(p=1\) diverges; convergence requires \(p>1\).

- Write the harmonic series as its first \(9\) terms plus the remaining tail. - The removed portion is a finite sum. - Decide whether a finite change can alter the convergence of an infinite series.

1. Removing the first \(9\) terms leaves the tail \(\sum_{n=10}^{\infty}\frac1n\). 2. The original harmonic series equals a finite prefix plus this tail. 3. Adding or removing finitely many terms does not change convergence, so the tail still diverges.

No. The remaining tail still diverges.

- Use the quotient rule for exponents to combine the powers of \(n\). - Rewrite a negative power as a reciprocal power to identify \(p\). - Compare the resulting exponent with the \(p\)-series boundary \(1\).

1. \(\frac{n^{2/3}}{n^2}=n^{2/3-2}=n^{-4/3}=\frac1{n^{4/3}}\). 2. Thus the effective exponent is \(p=\frac43\). 3. Since \(\frac43>1\), the \(p\)-series converges.

\(p=\frac43\), so the series converges.

- Replace \(\sqrt n\) with \(n^{1/2}\). - Subtract exponents when dividing powers with the same base. - Compare the resulting reciprocal-power exponent with \(1\).

1. \(\frac{\sqrt n}{n}=\frac{n^{1/2}}{n}=n^{-1/2}=\frac1{n^{1/2}}\). 2. Thus \(p=\frac12\). 3. Since \(\frac12\le1\), the series diverges.

\(p=\frac12\), so the series diverges.

- Subtract the denominator exponent from the numerator exponent. - Convert the resulting negative exponent to reciprocal form. - Compare the identified \(p\)-value with the convergence cutoff.

1. \(\frac{n^4}{n^{9/2}}=n^{4-9/2}=n^{-1/2}=\frac1{n^{1/2}}\). 2. Thus \(p=\frac12\). 3. Since \(\frac12\le1\), the series diverges.

\(p=\frac12\), so the series diverges.

- Rewrite the square root as a fractional power. - Add exponents when multiplying powers with the same base. - Compare the combined denominator exponent with \(1\).

1. \(n\sqrt n=n\cdot n^{1/2}=n^{3/2}\), so the summand is \(\frac1{n^{3/2}}\). 2. Thus \(p=\frac32\). 3. Since \(\frac32>1\), the \(p\)-series converges.

\(p=\frac32\), so the series converges.

- Factor the signed constant out of the series. - Classify the remaining reciprocal-power series by its exponent. - Multiplying every partial sum by a nonzero constant cannot turn a divergent series into a convergent one.

1. \(\sum_{n=1}^{\infty}-\frac3n=-3\sum_{n=1}^{\infty}\frac1n\), whose underlying \(p\)-series has \(p=1\). 2. The harmonic series diverges. 3. Multiplication by the nonzero constant \(-3\), including its negative sign, does not change convergence or divergence.

The underlying series has \(p=1\); the factor \(-3\) does not change divergence.

- Use \(\sqrt[k]{n^m}=n^{m/k}\) to rewrite the denominator. - Identify the resulting \(p\)-value. - Compare that exponent with the convergence boundary \(1\).

1. \(\sqrt[3]{n^5}=n^{5/3}\), so the summand is \(\frac1{n^{5/3}}\). 2. Thus \(p=\frac53\). 3. Since \(\frac53>1\), the \(p\)-series converges.

\(p=\frac53\), so the series converges.

- Rewrite the fourth root as a rational exponent. - Subtract exponents when dividing by \(n\), then convert the negative power to reciprocal form. - Compare the resulting \(p\)-value with \(1\).

1. \(\sqrt[4]{n^3}=n^{3/4}\), so \(\frac{n^{3/4}}n=n^{-1/4}=\frac1{n^{1/4}}\). 2. Thus \(p=\frac14\). 3. Since \(\frac14\le1\), the \(p\)-series diverges.

\(p=\frac14\), so the series diverges.

- Use the power-of-a-power rule on \((n^2)^{3/4}\). - Match the simplified denominator with \(n^p\). - Compare the resulting \(p\)-value with \(1\).

1. \((n^2)^{3/4}=n^{(2)(3/4)}=n^{3/2}\). 2. The series is \(\sum_{n=1}^{\infty}\frac1{n^{3/2}}\), so \(p=\frac32\). 3. Since \(\frac32>1\), the series converges.

The equivalent series is \(\sum_{n=1}^{\infty}\frac1{n^{3/2}}\); \(p=\frac32\), so it converges.

- Convert the square root to the exponent \(\frac12\). - Subtract exponents in the quotient and rewrite the negative power as a reciprocal. - Apply the \(p\)-series cutoff to the simplified exponent.

1. \(\sqrt{n^7}=n^{7/2}\), so \(\frac{n^2}{\sqrt{n^7}}=n^{2-7/2}=n^{-3/2}=\frac1{n^{3/2}}\). 2. Thus \(p=\frac32\). 3. Since \(\frac32>1\), the series converges.

\(p=\frac32\), so the series converges.

- Subtract the denominator exponent from the numerator exponent. - Convert the net negative exponent to standard reciprocal-power form. - State whether the resulting \(p\)-value lies above, below, or at \(1\).

1. \(\frac{n^{5/2}}{n^3}=n^{5/2-3}=n^{-1/2}=\frac1{n^{1/2}}\). 2. Thus \(p=\frac12\), which is below the boundary value \(1\). 3. Therefore the series diverges.

\(p=\frac12\), which is below the convergence boundary, so the series diverges.

- Rewrite the fifth root using the exponent \(\frac15\). - Combine the powers in the quotient and express the result as \(1/n^p\). - Compare \(p\) with the convergence boundary \(1\).

1. \(\sqrt[5]{n^3}=n^{3/5}\), so \(\frac{n^{3/5}}{n^2}=n^{3/5-2}=n^{-7/5}=\frac1{n^{7/5}}\). 2. Thus \(p=\frac75\). 3. Since \(\frac75>1\), the series converges.

\(p=\frac75\), so the series converges.

- Subtract the exponents in the quotient. - Rewrite the resulting negative power as a reciprocal power. - Recognize the special \(p=1\) boundary case.

1. \(\frac{n^{7/3}}{n^{10/3}}=n^{7/3-10/3}=n^{-1}=\frac1n\). 2. Thus \(p=1\), the harmonic boundary case. 3. Therefore the series diverges.

\(p=1\), so the series is harmonic and diverges.

- Combine the quotient of powers and rewrite it as \(1/n^p\). - Express the effective \(p\)-value in terms of \(r\). - Set that expression equal to the boundary value \(1\) and solve.

1. \(\frac{n^{3/2}}{n^r}=n^{3/2-r}=\frac1{n^{r-3/2}}\). 2. The effective exponent is \(p=r-\frac32\). 3. Setting \(p=1\) gives \(r-\frac32=1\), so \(r=\frac52\).

\(r=\frac52\).

- Subtract exponents in the quotient before identifying \(p\). - Rewrite the net power in reciprocal form. - Solve the inequality \(p>1\) for \(t\), including the boundary in the divergent range.

1. \(\frac{n^2}{n^{t+1}}=n^{1-t}=\frac1{n^{t-1}}\). 2. Thus \(p=t-1\), and convergence requires \(t-1>1\). 3. Therefore the series converges for \(t>2\) and diverges for \(t\le2\).

The series converges for \(t>2\) and diverges for \(t\le2\).

- Split the expression into two separate reciprocal-power series. - Classify each component by its own \(p\)-value. - Use the theorem that the sum of two convergent series converges.

1. \(\sum_{n=1}^{\infty}\frac1{n^2}\) is a convergent \(p\)-series with \(p=2\). 2. \(\sum_{n=1}^{\infty}\frac1{n^4}\) is a convergent \(p\)-series with \(p=4\). 3. The term-by-term sum of two convergent series also converges.

The series converges.

- Convert the square root to a fractional exponent before simplifying. - Subtract exponents in the quotient rather than discarding the numerator factor. - Apply the \(p\)-series criterion to the corrected exponent.

1. Rewrite \(\sqrt n\) as \(n^{1/2}\). 2. Then \(\frac{\sqrt n}{n^2}=n^{1/2-2}=n^{-3/2}=\frac1{n^{3/2}}\). 3. Since \(p=\frac32>1\), the series converges.

The correct form is \(\frac1{n^{3/2}}\), so \(p=\frac32\) and the series converges.

- Separate the full series according to index parity. - Factor the constant power from the even-indexed terms. - Subtract the even contribution from the full total.

1. The even-indexed subseries is \(\sum_{n=1}^{\infty}1/(2n)^p=2^{-p}S\). 2. The full series is the sum of its odd-indexed and even-indexed subseries. 3. Therefore the odd-indexed sum is \(S-2^{-p}S=(1-2^{-p})S\).

The odd-indexed terms sum to \((1-2^{-p})S\).

- Replace every occurrence of the sequence term by its power-law definition. - What must be true when two powers of the same positive integer agree for every index? - Use the recovered exponent to classify the original series.

1. Substituting \(a_n=n^{-p}\) gives \(n^{1-p}=n^{-p/2}\) for every \(n>1\). 2. Equal powers of the same variable must have equal exponents, so \(1-p=-p/2\). 3. Solving gives \(p=2\). 4. Therefore \(\sum a_n=\sum1/n^2\), which converges as a \(p\)-series.

\(p=2\), and the series converges.

- How does selecting only even indices change each term by a constant factor? - Can multiplication by that fixed factor alter convergence? - Make sure the argument covers every real exponent.

1. Factor the constant from the even-indexed subseries: \(\sum_{n=1}^{\infty}\frac{1}{(2n)^p}=2^{-p}\sum_{n=1}^{\infty}\frac{1}{n^p}\). 2. The factor \(2^{-p}\) is a finite positive constant for every real \(p\). 3. Multiplying a series by a nonzero constant does not change whether it converges. 4. Therefore the original \(p\)-series and its even-indexed subseries always have the same convergence behavior.

No. The two series always either both converge or both diverge.

- Identify the exact boundary, including whether equality is allowed. - Compare the given exponent with that boundary rather than rounding it. - Distinguish slow convergence from divergence.

1. A \(p\)-series converges exactly when \(p>1\). 2. Here \(p=1.0001>1\), so the series converges. 3. The exponent's numerical closeness to \(1\) can make convergence slow, but the classification uses the strict inequality \(p>1\). 4. Only the boundary value \(p=1\) gives the harmonic series.

The claim is false. The series converges because \(1.0001>1\).

- Bound the given power between nearby integer powers of the same base. - Use monotonicity of the exponential function to bound the exponent. - Only the exponent's position relative to the boundary is needed.

1. Since \(3^1=3<5<9=3^2\), the increasing function \(3^x\) gives \(1<p<2\). 2. In particular, \(p>1\). 3. Therefore the \(p\)-series converges.

The series converges.

- Translate the two given divergences into exponent bounds. - Average the bounds as well as the exponents. - Compare the average with the convergence boundary.

1. Divergence of the first two series gives \(p\le1\) and \(q\le1\). 2. Averaging these inequalities gives \(\frac{p+q}{2}\le1\). 3. Therefore the series with the average exponent also diverges.

No. The series with exponent \(\frac{p+q}{2}\) must diverge.

- Rewrite \(n^a\) as a reciprocal power so it matches \(1/n^p\). - Express the effective exponent \(p\) in terms of \(a\). - Solve the strict inequality \(p>1\), reversing signs carefully when needed.

1. Write \(n^a=\frac1{n^{-a}}\), so the effective \(p\)-value is \(p=-a\). 2. The series converges when \(-a>1\), which is equivalent to \(a<-1\). 3. It diverges when \(a\ge-1\).

The series converges for \(a<-1\) and diverges for \(a\ge-1\).

- Identify the exponent \(p\) as a linear expression in \(k\). - Impose the \(p\)-series convergence condition \(p>1\). - Solve the resulting inequality and include the boundary case in the divergent range.

1. The \(p\)-value is \(p=2k-1\). 2. Convergence requires \(2k-1>1\). 3. Thus the series converges for \(k>1\) and diverges for \(k\le1\).

The series converges for \(k>1\) and diverges for \(k\le1\).

- Combine the quotient of powers and rewrite it as \(1/n^p\). - Apply the strict inequality \(p>1\) to the expression \(6-m\). - Convert the resulting real inequality to the requested set of integer values.

1. \(\frac{n^m}{n^6}=n^{m-6}=\frac1{n^{6-m}}\), so \(p=6-m\). 2. Convergence requires \(6-m>1\), or \(m<5\). 3. Since \(m\) must be an integer, the solutions are all integers \(m\le4\).

All integers \(m\le4\).

- Subtract exponents in the quotient and convert the result to reciprocal-power form. - Identify \(p\) as a function of \(c\). - Solve \(p>1\) and place the equality case in the divergent range.

1. \(\frac{n^c}{n^4}=n^{c-4}=\frac1{n^{4-c}}\), so \(p=4-c\). 2. Convergence requires \(4-c>1\). 3. Thus the series converges for \(c<3\) and diverges for \(c\ge3\).

The series converges for \(c<3\) and diverges for \(c\ge3\).

- Identify the \(p\)-series exponent as \(5/q\). - Apply the strict \(p>1\) convergence condition. - Use the assumption \(q>0\) when solving the inequality.

1. The series has \(p=\frac5q\). 2. Convergence requires \(\frac5q>1\). 3. Because \(q>0\), multiplying by \(q\) preserves the inequality, giving \(q<5\). 4. Therefore the series converges for \(0<q<5\) and diverges for \(q\ge5\).

The series converges for \(0<q<5\) and diverges for \(q\ge5\).

- Write one inequality for convergence of the first \(p\)-series. - Write a separate inequality for divergence of the series with exponent \(p-1\). - Intersect the two solution sets and preserve the correct open and closed endpoints.

1. The first series converges when \(p>1\). 2. The second series diverges when \(p-1\le1\), which is equivalent to \(p\le2\). 3. Intersecting the two conditions gives \(1<p\le2\).

\(1<p\le2\).

- Separate the summand into a harmonic component and a \(p=2\) component. - Classify those two component series independently. - Use a contradiction: if the difference converged, what would happen after adding back the convergent component?

1. \(\sum_{n=1}^{\infty}\frac1n\) is the divergent harmonic series, while \(\sum_{n=1}^{\infty}\frac1{n^2}\) converges. 2. Suppose \(\sum_{n=1}^{\infty}\left(\frac1n-\frac1{n^2}\right)\) converged. 3. Adding the convergent series \(\sum_{n=1}^{\infty}\frac1{n^2}\) would then make \(\sum_{n=1}^{\infty}\frac1n\) converge, a contradiction. 4. Therefore the given series diverges.

The series diverges.

- Split the expression into a constant multiple of the harmonic series and a \(p=3\) series. - Classify each component before combining conclusions. - A convergent component cannot cancel the divergence of another component unless their divergent parts specifically offset, which they do not here.

1. \(\sum_{n=1}^{\infty}\frac4n\) is a nonzero constant multiple of the divergent harmonic series. 2. \(\sum_{n=1}^{\infty}\frac1{n^3}\) is a convergent \(p\)-series. 3. If the given sum converged, subtracting the convergent \(p=3\) series would force \(\sum_{n=1}^{\infty}\frac4n\) to converge, a contradiction. 4. Therefore the given series diverges.

The series diverges.

- Rewrite each radical as a rational exponent before comparing the series. - Order the three resulting \(p\)-values numerically. - Apply the \(p>1\) criterion to each exponent independently.

1. \(\frac1{\sqrt n}=\frac1{n^{1/2}}\), \(\frac1{n^2}\) has \(p=2\), and \(\frac1{\sqrt[3]{n^4}}=\frac1{n^{4/3}}\). 2. The increasing order of the exponents is \(\frac12<\frac43<2\). 3. The \(p=\frac12\) series diverges, while the \(p=\frac43\) and \(p=2\) series converge.

From smallest to largest \(p\)-value: \(\sum_{n=1}^{\infty}1/\sqrt n\) with \(p=\frac12\), \(\sum_{n=1}^{\infty}1/\sqrt[3]{n^4}\) with \(p=\frac43\), and \(\sum_{n=1}^{\infty}1/n^2\) with \(p=2\). The first diverges; the other two converge.

- One exponent must be at or below \(1\), and the other must be above \(1\). - Impose the additional requirement that the two exponents differ by exactly \(\frac12\). - Verify both the difference and the convergence classification of your chosen pair.

1. Choose two exponents that differ by \(\frac12\) and lie on opposite sides of the boundary \(p=1\), such as \(p=\frac34\) and \(p=\frac54\). 2. \(\sum_{n=1}^{\infty}1/n^{3/4}\) diverges and \(\sum_{n=1}^{\infty}1/n^{5/4}\) converges. 3. The exponents differ by \(\frac54-\frac34=\frac12\).

One valid pair is \(\sum_{n=1}^{\infty}1/n^{3/4}\), which diverges, and \(\sum_{n=1}^{\infty}1/n^{5/4}\), which converges.

- Simplify the ratio before using the given numerical condition. - Express the fixed scale factor and the fraction with a common base. - Compare the recovered exponent with the convergence boundary.

1. The ratio is \(\frac{a_{8n}}{a_n}=\frac{(8n)^{-p}}{n^{-p}}=8^{-p}\). 2. Thus \(8^{-p}=\frac14\). Writing both sides as powers of \(2\) gives \(2^{-3p}=2^{-2}\), so \(p=\frac23\). 3. Since \(p=\frac23\le1\), the corresponding \(p\)-series diverges.

\(p=\frac23\), and the series diverges.

- Rewrite the powers-of-two terms using the index \(k\). - The selected terms form a different standard type of series. - Intersect its convergence condition with the divergence condition for the full series.

1. The indexed subseries is \(\sum_{k=0}^{\infty}(2^{-p})^k\), a geometric series. 2. It converges when \(|2^{-p}|<1\), which is equivalent to \(p>0\). 3. The full \(p\)-series diverges when \(p\le1\). 4. Both conditions hold when \(0<p\le1\).

The indexed subseries converges for \(p>0\). It converges while the full \(p\)-series diverges for \(0<p\le1\).

- Determine how the extra factor changes the exponent. - Translate the requested behavior of each series into an inequality. - Intersect the two parameter ranges.

1. The original series converges when \(p>1\). 2. Since \(na_n=\frac{1}{n^{p-1}}\), the weighted series diverges when \(p-1\le1\), or \(p\le2\). 3. Combining the conditions gives \(1<p\le2\).

\(1<p\le2\).

- How do a positive number and its reciprocal compare on either side of the boundary value? - Classify the two exponents in separate parameter regions. - Treat the boundary case on its own.

1. If \(0<p<1\), then \(p\le1\) while \(1/p>1\). The first series diverges and the second converges. 2. If \(p=1\), both exponents equal \(1\), so both series diverge. 3. If \(p>1\), then \(p>1\) while \(1/p<1\). The first series converges and the second diverges.

\(0<p<1\): first diverges, second converges. \(p=1\): both diverge. \(p>1\): first converges, second diverges.

- Simplify the quotient by subtracting exponents. - Identify the effective exponent in standard reciprocal-power form. - Compare that exponent with the convergence boundary.

1. The quotient is \(\frac{a_n}{b_n}=\frac{n^{-p}}{n^{-q}}=n^{q-p}=\frac{1}{n^{p-q}}\). 2. This is a \(p\)-series with effective exponent \(p-q\). 3. It converges exactly when \(p-q>1\).

The series converges exactly when \(p-q>1\).

- Rewrite the quadratic exponent by completing the square. - Compare the smallest possible exponent with the convergence boundary. - Check separately when the squared expression becomes zero.

1. Rewrite the exponent as \(k^2-4k+5=(k-2)^2+1\). 2. The series converges when \((k-2)^2+1>1\), which is equivalent to \((k-2)^2>0\). 3. Thus the series converges for \(k\ne2\). 4. At \(k=2\), the exponent is \(1\), so the series is harmonic and diverges.

The series converges for \(k\ne2\) and diverges for \(k=2\).

- Convert each convergence statement into an inequality for the integer. - Keep the strict and non-strict boundary conditions distinct. - Use the integer restriction only after finding the interval.

1. Divergence of the first series requires \(m/3\le1\), so \(m\le3\). 2. Convergence of the second requires \((m+1)/3>1\), so \(m>2\). 3. The only integer satisfying \(2<m\le3\) is \(m=3\).

\(m=3\).

- Compare each finite-parameter exponent with the boundary. - Find the limit of the exponents separately from classifying the individual series. - Treat the limiting exponent as a new series rather than as a completed infinite list of conclusions.

1. For every positive integer \(m\), the exponent \(1+1/m\) is greater than \(1\), so \(S_m\) converges. 2. As \(m\to\infty\), the exponent approaches \(1\). 3. Replacing the exponent by its limit gives \(\sum_{n=1}^{\infty}\frac1n\), the harmonic series, which diverges. 4. Convergence for every member of a parameterized family does not imply convergence of the boundary series obtained from the limiting parameter.

Every \(S_m\) converges, but the limiting-exponent series \(\sum 1/n\) diverges.

- Expand a finite partial sum rather than separating two infinite series. - Track the endpoint terms after cancellation. - Divergent component series cannot always be subtracted individually.

1. The \(N\)th partial sum telescopes: \(S_N=1-1/(N+1)^p\). 2. Since \(p>0\), \(1/(N+1)^p\to0\). 3. Therefore the series converges to \(1\). 4. When \(0<p\le1\), the two separate \(p\)-series diverge, but their termwise difference has exact cancellation in every finite partial sum.

The series converges to \(1\) for every \(p>0\).

- Compare the rational exponent directly with the boundary value. - Move the boundary to one side and simplify to a sign question. - Track the excluded value and the equality case separately.

1. The series converges when \(\frac{2t+3}{t+2}>1\). 2. Subtracting \(1\) gives \(\frac{t+1}{t+2}>0\). 3. This quotient is positive when numerator and denominator have the same sign, so \(t<-2\) or \(t>-1\). 4. The series diverges for \(-2<t\le-1\); the expression is undefined at \(t=-2\).

The series converges for \(t<-2\) or \(t>-1\), diverges for \(-2<t\le-1\), and is undefined at \(t=-2\).

- Rewrite each selected subseries using the new index. - Compare the two effective exponents with the same boundary. - Account for the fact that one convergence condition is stronger than the other.

1. The square-indexed subseries is a \(p\)-series with exponent \(2p\), so it converges when \(p>1/2\). 2. The cube-indexed subseries has exponent \(3p\), so it converges when \(p>1/3\). 3. The cube-indexed series can converge while the square-indexed series diverges when \(1/3<p\le1/2\). 4. The reverse situation is impossible because \(p>1/2\) also implies \(p>1/3\).

Exactly one converges for \(\frac13<p\le\frac12\); in that interval only the cube-indexed subseries converges.

- Write the convergence interval for each exponent separately. - Mark the overlap where both series converge. - Keep the boundary values with the divergent series.

1. The first series converges when \(p>1\). 2. The second converges when \(3-p>1\), or \(p<2\). 3. If \(p\le1\), only the second series converges. If \(p\ge2\), only the first series converges. 4. For \(1<p<2\), both series converge.

Exactly one series converges when \(p\le1\) or \(p\ge2\).

- How does restricting the index to perfect squares change the effective exponent? - Write the two convergence requirements separately. - Intersect the resulting parameter ranges.

1. The full series diverges when \(p\le1\). 2. The square-indexed subseries is \(\sum_{k=1}^{\infty}1/k^{2p}\), which converges when \(2p>1\). 3. Combining \(p\le1\) with \(p>1/2\) gives \(1/2<p\le1\).

\(\frac12<p\le1\).

- How many terms occur in each block between successive powers of the same base? - What uniform lower bound applies to every term in a block? - What happens when infinitely many blocks each contribute a fixed positive amount?

1. For \(k\ge1\), the block after \(1\) contains the terms with \(2^{k-1}<n\le2^k\). 2. This block has \(2^{k-1}\) terms, and every term is at least \(1/2^k\). 3. Therefore each block has sum at least \(2^{k-1}\cdot\frac{1}{2^k}=\frac12\). 4. After \(m\) such blocks, the partial sum is at least \(1+\frac{m}{2}\), which grows without bound. Hence the harmonic series diverges.

Every block contributes at least \(\frac12\), so the partial sums are unbounded and the harmonic series diverges.

- Treat the two positive components separately. - Write one condition for each exponent. - Intersect the resulting intervals.

1. A sum of two positive series converges exactly when both component series converge. 2. The first component converges when \(p>1\). 3. The second component converges when \(4-p>1\), or \(p<3\). 4. Therefore the given series converges for \(1<p<3\) and diverges otherwise.

The series converges for \(1<p<3\) and diverges otherwise.

- Find the smallest possible value of the trigonometric contribution. - Determine when that minimum is attained. - Separate the boundary exponent from all larger exponents.

1. The exponent \(1+\sin^2\theta\) is always at least \(1\). 2. It equals \(1\) exactly when \(\sin\theta=0\), which occurs for \(\theta=k\pi\), where \(k\) is an integer. 3. For \(\theta=k\pi\), the series is harmonic and diverges. 4. For every other \(\theta\), the exponent is greater than \(1\), so the series converges.

The series diverges when \(\theta=k\pi\) for an integer \(k\), and converges otherwise.

- What condition must the reciprocal exponent satisfy for convergence? - Translate that condition into a distance from the excluded parameter value. - Keep the point where the expression is undefined separate from the intervals.

1. The series converges when \(\frac{1}{|t-2|}>1\). 2. Since \(|t-2|>0\), this is equivalent to \(|t-2|<1\). 3. Thus \(1<t<3\), with \(t=2\) excluded because the exponent is undefined. 4. The series diverges for \(t\le1\) or \(t\ge3\).

The series converges for \(1<t<3\) with \(t\ne2\), diverges for \(t\le1\) or \(t\ge3\), and is undefined at \(t=2\).

- Express the transformed terms using one effective exponent. - Write the convergence requirement for each series separately. - Do the resulting parameter conditions overlap?

1. The first series converges only when \(p>1\). 2. Since \(n^2a_n^3=\frac{n^2}{n^{3p}}=\frac{1}{n^{3p-2}}\), the second series diverges when \(3p-2\le1\). 3. The second condition simplifies to \(p\le1\), which contradicts \(p>1\). 4. Therefore no real value of \(p\) satisfies both requirements.

No such real value of \(p\) exists.

- Write one convergence condition for each original exponent. - Determine the exponent created by multiplying corresponding terms. - Describe the intersection of all three inequalities.

1. The first two series diverge when \(p\le1\) and \(q\le1\), respectively. 2. The product of corresponding terms is \(\frac{1}{n^{p+q}}\). 3. The product-term series converges when \(p+q>1\). 4. Therefore the required region is \(p\le1\), \(q\le1\), and \(p+q>1\).

All ordered pairs satisfying \(p\le1\), \(q\le1\), and \(p+q>1\).

- Translate each term transformation into a new exponent. - Write one inequality for each requested convergence behavior. - Intersect all three parameter ranges.

1. The original series diverges when \(p\le1\). 2. The squared-term series converges when \(2p>1\), so \(p>\frac12\). 3. Since \(na_n^3=\frac{1}{n^{3p-1}}\), the third series converges when \(3p-1>1\), so \(p>\frac23\). 4. Combining all conditions gives \(\frac23<p\le1\).

\(\frac23<p\le1\).

- Order the three exponents before checking convergence. - Identify which two must be above the boundary if exactly two converge. - Keep the equality case with the divergent side.

1. The exponents are ordered from smallest to largest as \(p-1<p+\frac12<p+2\). 2. Exactly two series converge when the two larger exponents exceed \(1\) and the smallest does not. 3. Thus \(p+\frac12>1\) and \(p-1\le1\). 4. These inequalities give \(\frac12<p\le2\).

\(\frac12<p\le2\).

- For indices greater than one, compare powers by comparing their exponents. - Determine which exponent controls each pointwise operation. - Apply the boundary criterion to the resulting effective exponents.

1. For \(n>1\), the smaller term is the one with the larger exponent. Thus the minimum equals \(1/n^{\max(p,q)}\). 2. The minimum series converges exactly when \(\max(p,q)>1\). 3. The larger term has the smaller exponent, so the maximum equals \(1/n^{\min(p,q)}\). 4. The maximum series converges exactly when \(\min(p,q)>1\), equivalently when both \(p>1\) and \(q>1\).

The minimum series converges iff \(\max(p,q)>1\). The maximum series converges iff \(\min(p,q)>1\).

- Express the consecutive-term ratio using a small reciprocal input. - Interpret the resulting quotient as a derivative at zero. - Use the recovered exponent in the standard classification.

1. The ratio is \(a_{n+1}/a_n=(n/(n+1))^p=(1+1/n)^{-p}\). 2. With \(x=1/n\), the limit becomes \(\lim_{x\to0^+}\frac{1-(1+x)^{-p}}{x}\). 3. This is the negative derivative of \((1+x)^{-p}\) at \(0\), so the limit is \(p\). 4. Therefore \(p=5/2\), and the \(p\)-series converges because \(5/2>1\).

\(p=\frac52\), and the series converges.

- Can the exponent be written using one positive variable and its reciprocal? - What is the smallest possible value of that symmetric expression? - Identify separately where equality occurs and where it is strict.

1. The exponent can be written as \(\frac12\left(|t|+\frac{1}{|t|}\right)\). 2. For \(|t|>0\), this quantity is at least \(1\), with equality only when \(|t|=1\). 3. Therefore the exponent exceeds \(1\) for every \(t\ne0,\pm1\), so the series converges there. 4. At \(t=\pm1\), the exponent is \(1\), so the series diverges. The expression is undefined at \(t=0\).

The series converges for all real \(t\ne0,\pm1\), diverges for \(t=\pm1\), and is undefined at \(t=0\).

- The smaller root controls whether both exponents exceed the boundary. - First impose the condition for real roots and a positive sum. - Isolate the radical carefully before squaring the inequality.

1. Real roots require \(t^2-8\ge0\). For both roots to exceed \(1\), their positive sum forces \(t\ge2\sqrt2\). 2. The smaller root is \(\frac{t-\sqrt{t^2-8}}{2}\). Both series converge exactly when this root is greater than \(1\). 3. For \(t\ge2\sqrt2\), the inequality \(t-\sqrt{t^2-8}>2\) is equivalent to \(t<3\). 4. Therefore \(2\sqrt2\le t<3\).

\(2\sqrt2\le t<3\).