- Think about partial sums of nonnegative terms and how an upper bound affects them. - Now consider what a lower bound does to partial sums that grow without bound. - Decide whether the inequalities have to hold at every index or only after some point.

1. If \(0\le a_n\le b_n\) eventually and \(\sum b_n\) converges, then \(\sum a_n\) converges. 2. If \(0\le b_n\le a_n\) eventually and \(\sum b_n\) diverges, then \(\sum a_n\) diverges.

Smaller than a convergent series implies convergence; larger than a divergent series implies divergence.

- Compare the denominators before comparing the fractions. - Think about the size of reciprocals of positive numbers.

1. Because \(n+4>n>0\), taking reciprocals reverses the inequality. 2. Thus \(1/(n+4)<1/n\) for every positive integer \(n\).

\(\frac1n>\frac1{n+4}\), because its denominator is smaller and positive.

- Which series has the larger term at every index? - What can a convergent upper comparison tell you about a positive-term series?

1. Each term of the given series is positive and no larger than the corresponding term of \(\sum_{n=1}^{\infty}\frac{1}{2^n}\). 2. The upper geometric series converges, so the given series converges by direct comparison.

The series converges by direct comparison with the convergent geometric series \(\sum_{n=1}^{\infty}\frac{1}{2^n}\).

- Which series has the smaller term at every index? - What does a divergent lower comparison say about the larger positive-term series?

1. The given terms are at least \(\frac{2}{n}>0\). 2. Since \(\sum_{n=1}^{\infty}\frac{2}{n}\) is a divergent multiple of the harmonic series, the given series diverges by direct comparison.

The series diverges by direct comparison with \(\sum_{n=1}^{\infty}\frac{2}{n}\).

- Which simpler positive series has comparable long-term size? - Find a term-by-term inequality, not just a limit of ratios. - Does the benchmark's behavior transfer in the direction you need?

1. Compare with \(\sum \frac1{n^2}\), which converges. 2. For every \(n\ge1\), \(0<\frac1{n^2+5}\le\frac1{n^2}\). 3. Therefore the given series converges by the direct comparison test.

\(0<\frac1{n^2+5}\le\frac1{n^2}\); the upper \(p=2\) series converges, so the given series converges by direct comparison.

- Compare \(n^3+1\) with \(n^3\) to create an upper bound for the positive summand. - Simplify the resulting bound to a reciprocal power of \(n\). - For a convergence proof, the upper benchmark must itself be a convergent series.

1. Compare with \(\sum \frac1{n^2}\), which converges. 2. For \(n\ge1\), \(n^3+1\ge n^3\), so \(\frac{n}{n^3+1}\le\frac1{n^2}\). 3. Therefore the given series converges by the direct comparison test.

For \(n\ge1\), \(0<\frac{n}{n^3+1}\le\frac1{n^2}\). Since \(\sum_{n=1}^{\infty}1/n^2\) converges, the given series converges by direct comparison.

- The leading quotient \(n/n^2\) suggests using the harmonic term \(1/n\). - Form the ratio of the given term to that benchmark and simplify before taking the limit. - A positive finite ratio transfers the harmonic series’ divergence.

1. Compare with \(\sum \frac1n\), which diverges. 2. \(\lim_{n\to\infty}\frac{n/(n^2+1)}{1/n}=\lim\frac{n^2}{n^2+1}=1\). 3. Therefore the given series diverges by the limit comparison test.

Using \(b_n=1/n\), \(\lim_{n\to\infty}\frac{n/(n^2+1)}{1/n}=1\). Since the harmonic series diverges, the given series diverges by limit comparison.

- Compare the highest-degree term in the numerator with the highest-degree term in the denominator. - Choose the reciprocal-power benchmark suggested by that dominant quotient. - Compute the ratio limit and use its positive finite value to transfer the benchmark’s behavior.

1. The leading terms give \(\frac{2n}{n^3}=\frac2{n^2}\), so use \(b_n=\frac1{n^2}\). 2. \(\lim_{n\to\infty}\frac{(2n+1)/(n^3+4)}{1/n^2}=\lim_{n\to\infty}\frac{2n^3+n^2}{n^3+4}=2\). 3. Because the limiting ratio is positive and finite and \(\sum 1/n^2\) converges, the given series converges by the limit comparison test.

The dominant reciprocal power is \(1/n^2\). The limiting ratio is \(2\), so the series converges.

- Compare the highest powers in the numerator and denominator to choose a reciprocal-power benchmark. - Multiply by the reciprocal of that benchmark and simplify the rational expression. - Use the positive finite ratio limit to transfer the benchmark’s behavior.

1. Compare with \(\sum \frac1n\), which diverges. 2. \(\lim_{n\to\infty}\frac{(n^2+3)/(n^3+1)}{1/n}=1\). 3. Therefore the given series diverges by the limit comparison test.

Using \(b_n=1/n\), \(\lim_{n\to\infty}\frac{(n^2+3)/(n^3+1)}{1/n}=1\). Since the harmonic series diverges, the given series diverges by limit comparison.

- Replace the denominator by the smaller quantity \(4^n\) to obtain an upper bound for the positive term. - Simplify that upper bound as a geometric term. - Check that the resulting common ratio has absolute value less than \(1\).

1. Compare with \(\sum \left(\frac34\right)^n\), which converges. 2. Since \(4^n+n\ge4^n\), \(\frac{3^n}{4^n+n}\le\left(\frac34\right)^n\). 3. Therefore the given series converges by the direct comparison test.

Since \(0<\frac{3^n}{4^n+n}\le(\frac34)^n\) and \(\sum_{n=1}^{\infty}(\frac34)^n\) converges, the given series converges by direct comparison.

- Use the stated bound on \(\sin^2 n\) before dividing by the positive quantity \(n^2\). - Identify the reciprocal-square series as an upper benchmark. - Apply direct comparison using the convergence of the \(p=2\) series.

1. Compare with \(\sum \frac1{n^2}\), which converges. 2. Because \(0\le\sin^2 n\le1\), \(0\le\frac{\sin^2 n}{n^2}\le\frac1{n^2}\). 3. Therefore the given series converges by the direct comparison test.

Because \(0\le\sin^2 n\le1\), \(0\le\frac{\sin^2 n}{n^2}\le\frac1{n^2}\). Since \(\sum_{n=1}^{\infty}1/n^2\) converges, the given series converges by direct comparison.

- Use the nonnegativity of \(|\sin n|\) to obtain a lower bound for the numerator. - Divide that bound by the positive denominator \(n\). - Compare with the harmonic series to prove divergence.

1. Compare with \(\sum \frac1n\), which diverges. 2. Since \(1+|\sin n|\ge1\), \(\frac{1+|\sin n|}{n}\ge\frac1n\). 3. Therefore the given series diverges by the direct comparison test.

Since \(1+|\sin n|\ge1\), \(\frac{1+|\sin n|}{n}\ge\frac1n\). The harmonic series diverges, so the given series diverges by direct comparison.

- Simplify the displayed ratio by multiplying the given term by \(n^2\). - Use leading powers to express the limit directly in terms of \(c\). - After matching the required limit, compare with the convergent \(p=2\) benchmark.

1. The limit equals \(c\). 2. Set \(c=3\). 3. The terms are then comparable to \(3/n^2\), so the series converges.

\(c=3\), and the series converges.

- Focus on the eventual inequality \(b_n\ge a_n>0\). - Think about whether the larger series could converge while the smaller one diverges. - Try arguing by contradiction using the convergence direction of direct comparison.

1. The larger terms satisfy \(b_n\ge a_n>0\). 2. A convergent \(\sum b_n\) would force the smaller \(\sum a_n\) to converge by comparison. 3. Therefore \(\sum b_n\) diverges.

\(\sum b_n\) diverges.

- Think about what an upper bound does to the partial sums of nonnegative terms. - Does the given inequality provide an upper bound or a lower bound on the series Milica is testing?

1. The given series is larger than a convergent series. 2. Being larger than a convergent series gives no convergence conclusion.

Being larger than a convergent series does not prove convergence. To prove convergence by direct comparison, Milica would need a convergent upper bound instead.

- Separate the first \(12\) terms from the remaining infinite tail. - A finite prefix cannot alter convergence or divergence. - Apply the comparison theorem only on the range where the inequality is valid.

1. Convergence is determined by the behavior of an infinite tail. 2. The first \(12\) terms form a finite sum and cannot change convergence. 3. Therefore the comparison test may be applied to the tail beginning at \(n=13\).

Yes. An eventual inequality is sufficient, so the comparison may begin at \(n=13\).

- Compare the highest powers in the numerator and denominator to find the net reciprocal power. - Choose the listed benchmark with that same order. - Compute the ratio limit and use the benchmark’s convergence classification.

1. The leading behavior is \(\frac{7n^3}{5n^5}=\frac7{5n^2}\). 2. Thus \(b_n=1/n^2\) has the same dominant order. 3. \(\lim_{n\to\infty}\frac{(7n^3+2)/(5n^5+n)}{1/n^2}=\frac75\). 4. Since the ratio limit is positive and finite and \(\sum1/n^2\) converges, the given series converges.

Use \(\sum_{n=1}^{\infty}1/n^2\). The limit-comparison constant is \(\frac75\), so the series converges.

- Check whether “smaller than a divergent series” is one of the valid direct-comparison directions. - Find an example where the smaller series converges while the larger one diverges. - Find another allowed example where both series diverge.

1. Being smaller than a divergent series is not a valid comparison direction. 2. The smaller series can converge: for example, \(1/n^2\le1/n\), while \(\sum1/n^2\) converges and \(\sum1/n\) diverges. 3. The smaller series can also diverge: take \(a_n=b_n=1/n\). 4. Therefore no conclusion follows about \(\sum a_n\).

No conclusion follows; \(\sum a_n\) may converge or diverge.

- Check whether the terms keep one nonnegative sign as \(n\) changes. - Recall the sign hypothesis for the standard direct and limit comparison tests. - If absolute convergence is the goal, apply comparison to the absolute values instead.

1. The direct and limit comparison tests in this setting require nonnegative terms. 2. The terms \((-1)^n/n^2\) change sign, so signed inequalities do not meet those hypotheses. 3. To study absolute convergence, one may instead compare \(|(-1)^n/n^2|=1/n^2\).

The signed terms are not nonnegative, so the standard positive-term comparison tests do not apply directly.

- How can you compare powers that have the same base but different exponents? - Look for a term-by-term inequality that holds at every positive integer index. - Which of the given series can be the upper comparison?

1. Since \(n\ge1\) and \(q>p\), \(n^q\ge n^p>0\). 2. Taking reciprocals gives \(0<\frac{1}{n^q}\le\frac{1}{n^p}\) for every \(n\ge1\). 3. The given upper comparison series converges, so direct comparison proves convergence of \(\sum_{n=1}^{\infty}\frac{1}{n^q}\).

For \(q>p\) and \(n\ge1\), \(0<\frac{1}{n^q}\le\frac{1}{n^p}\). The given upper series converges, so the second series converges by direct comparison.

- Bound the logarithm using a simpler expression involving its positive input. - Choose a standard benchmark with the same small-input order. - Use the comparison in the direction that proves convergence.

1. For \(x>0\), \(0<\ln(1+x)<x\). 2. With \(x=1/n^2\), \(0<\ln\!\left(1+\frac{1}{n^2}\right)<\frac{1}{n^2}\). 3. Since \(\sum 1/n^2\) converges, the given series converges by direct comparison.

\(0<\ln\!\left(1+\frac1{n^2}\right)<\frac1{n^2}\); the upper \(p=2\) series converges, so the given series converges.

- Bound the greatest-integer numerator by the expression inside it. - Simplify the resulting power of the index. - Compare with a standard positive series.

1. The floor function satisfies \(0\le\lfloor\sqrt n\rfloor\le\sqrt n\). 2. Therefore \(0\le\frac{\lfloor\sqrt n\rfloor}{n^2}\le\frac{\sqrt n}{n^2}=1/n^{3/2}\). 3. The comparison \(p\)-series converges because \(3/2>1\). 4. Hence the given series converges by direct comparison.

\(0\le\frac{\lfloor\sqrt n\rfloor}{n^2}\le\frac1{n^{3/2}}\). The upper \(p\)-series converges, so the given series converges.

- First combine the two benchmark series into one nonnegative benchmark. - Apply the comparison only on the tail where the inequality is known. - Account for the finitely many terms before that tail.

1. Since \(\sum b_n\) and \(\sum c_n\) converge, their termwise sum series \(\sum(b_n+c_n)\) also converges. 2. On a tail, \(0\le a_n\le b_n+c_n\). 3. The direct comparison test shows that the tail of \(\sum a_n\) converges. 4. Adding or removing finitely many initial terms does not affect convergence, so \(\sum a_n\) converges.

The series \(\sum a_n\) converges by comparison with the convergent series \(\sum(b_n+c_n)\).

- Separate the reciprocal-power factor from the remaining multiplier. - Use the reciprocal-power part as the benchmark. - Evaluate the multiplier's limit.

1. Choose \(b_n=1/n^{4/3}\), a convergent \(p\)-series. 2. The ratio is \(\frac{e^{-1/n}/n^{4/3}}{1/n^{4/3}}=e^{-1/n}\). 3. This ratio tends to \(1\). 4. Therefore the given series converges by limit comparison.

The series converges by limit comparison with \(\sum1/n^{4/3}\); the limiting ratio is \(1\).

- Keep the reciprocal linear factor as the benchmark. - Determine the limiting value of the remaining denominator factor. - Check that the ratio limit is positive and finite.

1. Let \(a_n=1/(n\arctan n)\) and \(b_n=1/n\). 2. The ratio is \(\frac{a_n}{b_n}=\frac{1}{\arctan n}\). 3. Since \(\arctan n\to\pi/2\), the ratio tends to \(2/\pi\), a positive finite constant. 4. The harmonic series diverges, so the given series diverges by limit comparison.

The series diverges by limit comparison with \(\sum1/n\); the limiting ratio is \(\frac{2}{\pi}\).

- 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.

- Which series has the smaller positive terms for sufficiently large indices? - Does multiplying a divergent positive series by a fixed positive constant change convergence? - Use the divergence direction of direct comparison.

1. On a tail, \(a_n\ge c b_n>0\). 2. Since \(c>0\), the series \(\sum c b_n\) diverges with \(\sum b_n\). 3. Direct comparison gives divergence of \(\sum a_n\).

\(\sum a_n\) diverges by comparison with the divergent positive series \(\sum c b_n\).

- Bound the alternating factor in the numerator before choosing a benchmark. - Look for a convergent positive series that is always at least as large as the given terms. - When comparing strategies, check whether the quotient needed for limit comparison actually has a limit.

1. Since \((-1)^n\) is either \(-1\) or \(1\), \(1\le2+(-1)^n\le3\). 2. Therefore \(0<\frac{2+(-1)^n}{n^3}\le\frac{3}{n^3}\). 3. The series \(\sum 3/n^3\) converges, so the given series converges by direct comparison. 4. Direct comparison is more efficient here. A limit comparison with \(1/n^3\) would require the ratio \(2+(-1)^n\), which has no limit.

The series converges by direct comparison with \(\sum 3/n^3\). Direct comparison is more efficient because the corresponding limit-comparison ratio oscillates.

- Compare the two positive denominators. - Use a familiar convergent reciprocal-power series as an upper bound. - Recall the direction in which a convergent upper bound is useful.

1. Since \(n^2+n\ge n^2\), we have \(0<1/(n^2+n)\le1/n^2\). 2. The series \(\sum1/n^2\) converges. 3. By direct comparison, the given series converges.

\(0<\frac1{n^2+n}\le\frac1{n^2}\). Since \(\sum1/n^2\) converges, the given series converges.

- Think of a simple divergent series that may serve as a lower bound. - Check which positive term is larger at each index. - Does your inequality work for every positive integer \(n\)?

1. Rewrite the term as \(\frac1n+\frac2{n^2}\). 2. Thus \(\frac{n+2}{n^2}\ge\frac1n>0\). 3. The harmonic series diverges, so the given larger positive-term series diverges by direct comparison.

\(\frac{n+2}{n^2}\ge\frac1n\). The harmonic series diverges, so the given series diverges.

- Form the quotient and simplify its powers before taking a limit. - Ask whether the result lies strictly between zero and infinity. - Use the known convergence of one series.

1. The ratio is \(a_n/b_n=\frac{3n^2}{n^2+1}\), which approaches \(3\) as \(n\to\infty\). 2. This is a positive finite limit, and \(\sum b_n\) converges because \(p=2\). 3. By limit comparison, \(\sum a_n\) converges.

The comparison limit is \(3\). Since \(\sum1/n^2\) converges, \(\sum\frac{3}{n^2+1}\) converges by limit comparison.

- What does a quotient tending to zero tell you about the relative sizes of the two terms? - Would an upper bound by a divergent series settle convergence of the smaller one? - What other fact about this specific series can settle the question without this zero-limit comparison?

1. The ratio is \((1/n^2)/(1/n)=1/n\to0\). 2. A zero ratio against a divergent benchmark is inconclusive, so this comparison alone does not prove convergence. 3. Independently, \(\sum1/n^2\) is a convergent \(p\)-series with \(p=2\).

No. The zero comparison limit against a divergent series is inconclusive. \(\sum1/n^2\) does converge by the \(p\)-series test.

- Think about which comparison direction proves convergence. - Can removing finitely many initial terms change convergence?

1. Being smaller than a divergent positive-term series does not prove convergence. 2. The given sum is a harmonic tail starting at \(1/2\), so it diverges.

A smaller series need not converge when the upper comparison diverges. The shifted harmonic series also diverges.

- Read the two curves at the same horizontal position. - Use the vertical scale to estimate each height at \(x=2\). - What happens to the argument when \(n\) lies beyond the displayed interval?

1. Curve \(g\) is higher at both \(x=1\) and \(x=2\). 2. At \(x=2\), the graph gives approximately \(f(2)=0.33\) and \(g(2)=0.50\). 3. The graph only covers a finite interval. It cannot prove the inequality for every later positive integer; that needs a general argument.

Curve \(g\) is higher at both inputs. At \(x=2\), \(f(2)\approx0.33\) and \(g(2)\approx0.50\). No finite graph proves the inequality for all positive integers.

- How does \(2+\sin n\) compare with a simple positive constant? - Find the smallest possible value of the oscillating numerator. - To prove divergence, seek a smaller divergent series.

1. Since \(-1\le\sin n\le1\), the numerator satisfies \(2+\sin n\ge1\) for all \(n\). 2. Therefore \(\frac{2+\sin n}{\sqrt n}\ge\frac1{\sqrt n}>0\). 3. The \(p=1/2\) benchmark diverges, so the given series diverges by direct comparison.

\(\frac{2+\sin n}{\sqrt n}\ge\frac1{\sqrt n}\). The lower \(p=\frac12\) series diverges, so the given series diverges.

- Which standard convergent series has similar-sized terms for large \(n\)? - What inequality would place the given positive terms below that benchmark, possibly after a constant multiplier? - Check that the inequality holds from the stated starting index.

1. Compare with \(\sum \frac2{n^2}\), which converges. 2. For \(n\ge3\), \(n^2-4\ge\frac12n^2\), so \(\frac1{n^2-4}\le\frac2{n^2}\). 3. Therefore the given series converges by the direct comparison test.

For \(n\ge3\), \(0<\frac1{n^2-4}\le\frac2{n^2}\). Since \(\sum 2/n^2\) converges, the series converges.

- Which familiar divergent series has terms of a similar size? - Could a simple upper bound on the denominator give a lower bound on the term? - Check the direction of the inequality before using direct comparison.

1. Compare with \(\sum \frac1{2n}\), which diverges. 2. For \(n\ge1\), \(\sqrt n\le n\), so \(n+\sqrt n\le2n\) and \(\frac1{n+\sqrt n}\ge\frac1{2n}\). 3. Therefore the given series diverges by the direct comparison test.

\(\frac1{n+\sqrt n}\ge\frac1{2n}\) for \(n\ge1\). The lower comparison series \(\sum1/(2n)\) diverges, so the given series diverges.

- First verify that \(n-\sqrt n\) is positive for the stated indices. - Compare the denominator \(n-\sqrt n\) with \(n\); remember how a smaller positive denominator affects a reciprocal. - Use the harmonic series as a divergent lower benchmark.

1. Compare with \(\sum \frac1n\), which diverges. 2. For \(n\ge2\), \(0<n-\sqrt n<n\), so \(\frac1{n-\sqrt n}>\frac1n\). 3. Therefore the given series diverges by the direct comparison test.

For \(n\ge2\), \(0<n-\sqrt n<n\), so \(1/(n-\sqrt n)>1/n\). Since the harmonic series diverges, the given series diverges by direct comparison.

- Rewrite \(\sqrt n\) as \(n^{1/2}\) and compare the dominant powers in numerator and denominator. - Choose the reciprocal-power benchmark indicated by that exponent difference. - Simplify the ratio limit and use the benchmark’s \(p\)-series classification.

1. Compare with \(\sum \frac1{n^{3/2}}\), which converges. 2. \(\lim_{n\to\infty}\frac{\sqrt n/(n^2+1)}{1/n^{3/2}}=\lim\frac{n^2}{n^2+1}=1\). 3. Therefore the given series converges by the limit comparison test.

Using \(b_n=1/n^{3/2}\), \(\lim_{n\to\infty}\frac{\sqrt n/(n^2+1)}{1/n^{3/2}}=1\). Since \(\sum_{n=1}^{\infty}1/n^{3/2}\) converges, the given series converges by limit comparison.

- Factor \(n^2\) from the expression under the square root to reveal the dominant \(1/n\) behavior. - Compare the term with the harmonic benchmark by forming their ratio. - A positive finite ratio transfers the harmonic series’ divergence.

1. Compare with \(\sum \frac1n\), which diverges. 2. \(\lim_{n\to\infty}\frac{1/\sqrt{n^2+n}}{1/n}=\lim\frac{n}{\sqrt{n^2+n}}=1\). 3. Therefore the given series diverges by the limit comparison test.

Using \(b_n=1/n\), \(\lim_{n\to\infty}\frac{1/\sqrt{n^2+n}}{1/n}=1\). Since the harmonic series diverges, the given series diverges by limit comparison.

- Write \(n!\) as \(1\cdot2\cdot3\cdots n\) and bound each factor after the first from below. - Reverse the resulting positive-product inequality when taking reciprocals. - Compare with a geometric series whose ratio is less than \(1\).

1. Compare with \(\sum \frac1{2^{n-1}}\), which converges. 2. For \(n\ge1\), \(n!=1\cdot2\cdots n\ge2^{n-1}\), so \(\frac1{n!}\le\frac1{2^{n-1}}\). 3. Therefore the given series converges by the direct comparison test.

For \(n\ge1\), \(n!\ge2^{n-1}\), so \(0<1/n!\le1/2^{n-1}\). The geometric benchmark converges, so \(\sum_{n=1}^{\infty}1/n!\) converges by direct comparison.

- Use the lower bound \(\sin n\ge-1\) to estimate the denominator on a suitable tail. - Compare \(n^2-1\) with a fixed positive fraction of \(n^2\). - Convert the denominator estimate into an upper \(p=2\) benchmark for convergence.

1. Compare with \(\sum \frac2{n^2}\), which converges. 2. For \(n\ge2\), \(n^2+\sin n\ge n^2-1\ge\frac12n^2\), so the term is at most \(\frac2{n^2}\). 3. Therefore the given series converges by the direct comparison test.

For \(n\ge2\), \(n^2+\sin n\ge n^2-1\ge\frac12n^2\), so \(0<\frac1{n^2+\sin n}\le\frac2{n^2}\). The \(p=2\) benchmark converges, so the given series converges.

- Use the upper bound \(\sin n\le1\) to estimate the denominator from above. - Translate that denominator bound into a lower bound for the reciprocal term. - Compare with the shifted harmonic series, whose divergence is unchanged by its starting index.

1. Compare with \(\sum \frac1{n+1}\), which diverges. 2. Since \(\sin n\le1\), \(n+\sin n\le n+1\), so \(\frac1{n+\sin n}\ge\frac1{n+1}\). 3. Therefore the given series diverges by the direct comparison test.

Since \(n+\sin n\le n+1\), \(\frac1{n+\sin n}\ge\frac1{n+1}\). The shifted harmonic benchmark diverges, so the given series diverges by direct comparison.

- The square root in the numerator has leading size \(n\), suggesting a harmonic benchmark. - Form the ratio to \(1/n\) and factor \(n^2\) inside the square root. - Use the positive finite ratio limit to transfer divergence.

1. Compare with \(\sum \frac1n\), which diverges. 2. \(\lim_{n\to\infty}\frac{\sqrt{n^2+1}/n^2}{1/n}=\lim\sqrt{1+\frac1{n^2}}=1\). 3. Therefore the given series diverges by the limit comparison test.

Using \(b_n=1/n\), \(\lim_{n\to\infty}\frac{\sqrt{n^2+1}/n^2}{1/n}=1\). Since the harmonic series diverges, the given series diverges by limit comparison.

- Multiply the difference of square roots by its conjugate to rewrite the summand. - Use the resulting denominator to identify a reciprocal-square-root benchmark. - Compute the ratio limit and transfer the divergent benchmark’s behavior.

1. Rationalizing gives \(\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}\). 2. Compare with \(b_n=\frac1{\sqrt n}\), whose \(p\)-series diverges because \(p=\frac12\). 3. \(\lim_{n\to\infty}\frac{1/(\sqrt{n+1}+\sqrt n)}{1/\sqrt n}=\lim_{n\to\infty}\frac{\sqrt n}{\sqrt{n+1}+\sqrt n}=\frac12\). 4. Because the limiting ratio is positive and finite, the given series diverges by the limit comparison test.

The summand is \(\frac1{\sqrt{n+1}+\sqrt n}\). Comparing with \(1/\sqrt n\) gives a limiting ratio of \(\frac12\), so the series diverges.

- The degree difference between numerator and denominator suggests a reciprocal-square benchmark. - Form the ratio to \(1/n^2\) and retain the leading coefficients \(a\) and \(c\). - Use the positive finite ratio together with convergence of the \(p=2\) series.

1. Compare with \(\sum1/n^2\). 2. \(\lim_{n\to\infty}\frac{(an+b)/(cn^3+d)}{1/n^2}=\frac ac\). 3. Because the limit is positive and finite and \(\sum1/n^2\) converges, the given series converges.

Using \(b_n=1/n^2\), the ratio limit is \(a/c>0\). Since \(\sum_{n=1}^{\infty}1/n^2\) converges, the given series converges by limit comparison.

- Compare the degrees of the leading numerator and denominator terms to select \(1/n^2\). - Multiply the given term by \(n^2\) and take the ratio limit using leading coefficients. - Transfer convergence from the reciprocal-square benchmark when the limit is positive and finite.

1. Use \(b_n=1/n^2\). 2. \(\lim_{n\to\infty}\frac{(5n^2+1)/(2n^4+3)}{1/n^2}=\frac52\). 3. Since \(\sum1/n^2\) converges, the given series converges.

Compare with \(\sum_{n=1}^{\infty}1/n^2\). The limit-comparison constant is \(\frac52\), so the series converges.

- Begin with a positive term whose dominant part is \(4/n^2\). - You may add a smaller positive-order term without changing the desired ratio limit. - Verify the ratio directly, then transfer convergence from the \(p=2\) benchmark.

1. One choice is \(a_n=\frac{4n+1}{n^3}\). 2. \(\frac{a_n}{1/n^2}=4+\frac1n\to4\). 3. Since \(\sum1/n^2\) converges, \(\sum a_n\) converges by the limit comparison test.

One example is \(\sum\frac{4n+1}{n^3}\), which converges.

- Identify why an upper bound by a divergent series does not transfer divergence. - For a convergence proof, seek an upper benchmark that is known to converge. - Compare \(n^2+n\) with \(n^2\) and reverse the positive-denominator inequality correctly.

1. The comparison is with a larger divergent series. 2. Being smaller than a divergent series gives no conclusion. 3. A valid comparison is \(\frac1{n^2+n}\le\frac1{n^2}\), which proves convergence.

The stated comparison is inconclusive; the series actually converges.

- Recall the precise range of ratio limits that gives equivalence in the standard limit comparison test. - Interpret a ratio limit of \(0\) as saying that \(a_n\) is much smaller than \(b_n\). - Ask whether being smaller than a divergent series is a valid direct-comparison direction.

1. The standard equivalence conclusion requires a ratio limit \(L\) with \(0<L<\infty\). 2. A limit of \(0\) means \(a_n\) is asymptotically smaller than \(b_n\). 3. A series smaller than a divergent benchmark may either converge or diverge, so this information is inconclusive.

No. The standard limit comparison test is inconclusive in this situation.

- Find a simple lower bound for the numerator on the stated tail. - Find a simple upper bound for the denominator on the same tail. - Compare from below with a known divergent series.

1. For \(n\ge3\), \(\ln n\ge1\) and \(\ln n\le n\). 2. Therefore \(n+\ln n\le2n\), so \(\frac{\ln n}{n+\ln n}\ge\frac{1}{2n}\). 3. Since \(\sum 1/(2n)\) diverges, the given series diverges by direct comparison.

\(\frac{\ln n}{n+\ln n}\ge\frac1{2n}\) for \(n\ge3\); the lower harmonic multiple diverges, so the series diverges.

- Recall which inequality direction transfers convergence. - Find a lower bound for the denominator using a fixed fraction of its leading term. - Convert that denominator bound into an upper bound for the terms.

1. Being larger than a convergent positive series does not imply convergence, so Rami's comparison is invalid. 2. For \(n\ge2\), \(n^2-n\ge n^2/2\). 3. Therefore \(0<1/(n^2-n)\le2/n^2\). 4. Since \(\sum2/n^2\) converges, the given series converges by direct comparison.

Rami's comparison is invalid because a convergent lower bound proves nothing about the larger series. But \(0<\frac1{n^2-n}\le\frac2{n^2}\) for \(n\ge2\), so the series converges.

- Compare the dominant powers before computing any ratio. - Test whether each proposed benchmark produces a finite nonzero limiting quotient. - Use the classification of the matching benchmark.

1. The leading-power quotient is \(n^3/n^{9/2}=1/n^{3/2}\), so choose \(b_n=1/n^{3/2}\). 2. Then \(a_n/b_n=\frac{n^{9/2}+n^{3/2}}{n^{9/2}+n^{1/2}}\to1\). 3. With \(1/n\), the ratio tends to \(0\); with \(1/n^2\), it tends to \(\infty\). Neither gives the standard finite-positive limit-comparison conclusion. 4. Since \(\sum1/n^{3/2}\) converges, \(\sum a_n\) converges.

The correct benchmark is \(1/n^{3/2}\), the limiting ratio is \(1\), and the series converges.

- Bound the entire finite numerator without evaluating it exactly. - Use the infinite version of the inner pattern as a uniform bound. - Compare the resulting outer terms with a standard series.

1. The finite geometric sum in the numerator is positive and less than the infinite sum \(1+1/2+1/4+\cdots=2\). 2. Therefore \(0<\frac{1+1/2+\cdots+1/2^n}{n^2}<\frac{2}{n^2}\). 3. The comparison series \(\sum2/n^2\) converges. 4. Hence the given series converges by direct comparison.

\(0<\frac{1+\frac12+\cdots+\frac1{2^n}}{n^2}<\frac2{n^2}\). The geometric numerator is below \(2\), and the upper \(p=2\) series converges, so the given series converges.

- Pair the small logarithmic increment with a reciprocal benchmark. - Rewrite the ratio using the input of the logarithm. - Use the standard small-input logarithmic limit.

1. Let \(a_n=\ln(1+1/n)\) and \(b_n=1/n\). 2. The ratio is \(\frac{a_n}{b_n}=n\ln\!\left(1+\frac1n\right)\). 3. Using \(\lim_{x\to0}\frac{\ln(1+x)}{x}=1\), the ratio tends to \(1\). 4. Since \(\sum b_n\) diverges, the given series diverges by limit comparison.

The series diverges; the limiting ratio with \(1/n\) is \(1\).

- Separate the sparse exceptional indices from the remaining indices. - Reindex the exceptional terms by their exponent of two. - Compare or classify each nonnegative subseries separately.

1. The nonspecial terms satisfy \(0\le a_n\le1/n^2\), so their subseries converges. 2. At \(n=2^k\), the special terms are \(1/\sqrt{2^k}=(1/\sqrt2)^k\). 3. The special-term subseries is geometric with ratio \(1/\sqrt2<1\), so it converges. 4. The full positive series is the sum of two convergent subseries and therefore converges.

The series converges.

- Relate an index in a block to the left endpoint of that block. - Compare the block's constant value with a familiar reciprocal term. - Use a lower comparison to transfer divergence.

1. If \(2^k\le n<2^{k+1}\), then \(n\ge2^k\), so \(a_n=1/2^k\ge1/n\). 2. Thus \(a_n\ge1/n\) for every positive integer \(n\). 3. The harmonic series diverges. 4. Therefore \(\sum a_n\) diverges by direct comparison.

\(a_n=2^{-k}\ge1/n\) whenever \(2^k\le n<2^{k+1}\). Comparison with the divergent harmonic series proves divergence.

- Estimate the leading size of the two radicals together. - Which reciprocal power matches the full term after division by the denominator? - Normalize the radicals by their common scale before taking the limit.

1. The numerator has order \(\sqrt n\), so choose \(b_n=1/n^{3/2}\). 2. The ratio is \(\frac{(\sqrt{n+1}+\sqrt n)/n^2}{1/n^{3/2}}=\sqrt{1+\frac1n}+1\). 3. The ratio tends to \(2\). 4. Since \(\sum1/n^{3/2}\) converges, the given series converges by limit comparison.

The series converges by limit comparison with \(\sum1/n^{3/2}\); the limiting ratio is \(2\).

- Match the trigonometric expression to its first nonzero small-input order. - Rewrite the sequence ratio using one variable tending to zero. - Compare with a reciprocal power having that order.

1. The small-angle behavior suggests \(b_n=1/n^2\). 2. The ratio is \(\frac{1-\cos(1/n)}{1/n^2}=\frac{1-\cos x}{x^2}\) with \(x=1/n\). 3. As \(x\to0\), the ratio tends to \(1/2\). 4. Since \(\sum1/n^2\) converges, the given series converges by limit comparison.

The series converges by limit comparison with \(\sum1/n^2\); the limiting ratio is \(\frac12\).

- Match each inequality direction with the behavior of the known series. - Ask whether the unknown series is being bounded above or below. - For an unsupported implication, test it with familiar benchmark series.

1. a) The information is sufficient: direct comparison transfers convergence from the larger positive series to the smaller one. 2. b) The information is sufficient: direct comparison transfers divergence from the smaller positive series to the larger one. 3. c) The information is not sufficient. For example, with \(b_n=1/n\), choosing \(a_n=1/n\) gives divergence, while choosing \(a_n=1/n^2\) gives convergence. 4. Thus only the two standard comparison directions are valid.

a) Sufficient; \(\sum a_n\) converges. b) Sufficient; \(\sum a_n\) diverges. c) Insufficient; either behavior is possible.

- Start with a familiar convergent positive benchmark. - Can two positive convergent series have a quotient that keeps changing? - Distinguish a sufficient test condition from a necessary condition.

1. Let \(b_n=1/n^2\) and \(a_n=[2+(-1)^n]/n^2\). 2. Since \(1/n^2\le a_n\le3/n^2\), both \(\sum a_n\) and \(\sum b_n\) converge by comparison with the \(p\)-series \(\sum1/n^2\). 3. The ratio is \(a_n/b_n=2+(-1)^n\), which alternates between \(1\) and \(3\), so it has no limit. 4. The limit comparison test gives a sufficient condition when a positive finite ratio limit exists; it does not say that such a limit is necessary for two series to have the same convergence behavior.

One choice is \(b_n=1/n^2\) and \(a_n=[2+(-1)^n]/n^2\). Both series converge, but \(a_n/b_n=2+(-1)^n\) has no limit. This does not contradict limit comparison: a positive finite ratio limit is sufficient, not necessary.

- Inspect the terms at the exceptional indices before considering the rest. - Reindex those terms using the square root of the index. - A nonnegative series cannot converge if one of its subseries diverges.

1. At square indices \(n=k^2\), the terms are \(a_{k^2}=1/k\). 2. The square-indexed subseries is therefore the harmonic series \(\sum1/k\), which diverges. 3. All terms are nonnegative, so the partial sums of the full series dominate the partial sums of this divergent subseries. 4. Hence the full series diverges.

The series diverges because its square-indexed subseries is harmonic.

- Bound the oscillating denominator factor from above. - Check positivity before reversing the denominator inequality. - Compare from below with a harmonic multiple.

1. Since \(\sin n\le1\), the denominator satisfies \(n(2+\sin n)\le3n\). 2. Also \(2+\sin n\ge1\), so all terms are positive. 3. Therefore \(\frac{1}{n(2+\sin n)}\ge\frac{1}{3n}\). 4. Since \(\sum1/(3n)\) diverges, the given series diverges by direct comparison.

\(\frac1{n(2+\sin n)}\ge\frac1{3n}\) because \(1\le2+\sin n\le3\). The harmonic lower bound diverges, so the series diverges.

- Use the larger denominator exponent to determine the leading scale. - Subtract exponents to select a reciprocal-power benchmark. - Normalize the ratio by the dominant denominator power.

1. The dominant denominator power is \(n^{5/6}\), so the summand has order \(1/n^{1/2}\). Choose \(b_n=1/\sqrt n\). 2. The ratio is \(\frac{n^{1/3}/(n^{5/6}+n^{1/2})}{1/n^{1/2}}=\frac{n^{5/6}}{n^{5/6}+n^{1/2}}\). 3. Dividing by \(n^{5/6}\) gives a limit of \(1\). 4. Since \(\sum1/\sqrt n\) diverges, the given series diverges by limit comparison.

The series diverges by limit comparison with \(\sum1/\sqrt n\); the limiting ratio is \(1\).

- Determine which denominator term eventually supplies a universal positive lower bound. - Use that bound to create an upper comparison for the reciprocal. - Initial indices do not affect convergence.

1. The denominator satisfies \(n^p+3^n\ge3^n\) whenever the expression is positive; for sufficiently large \(n\), it is positive for every real \(p\). 2. Therefore on a tail, \(0<1/(n^p+3^n)\le1/3^n\). 3. The geometric comparison series converges. 4. Thus the given series converges for every real \(p\).

The series converges for every real \(p\).

- Convert the zero limit into an eventual numerical bound for the ratio. - Use that bound to compare the terms directly. - Treat finitely many initial terms separately from the tail.

1. Since \(a_n/b_n\to0\), there is an \(N\) such that \(a_n/b_n<1\) for every \(n\ge N\). 2. Thus \(0<a_n<b_n\) on that tail. 3. The tail \(\sum_{n=N}^{\infty}b_n\) converges, so \(\sum_{n=N}^{\infty}a_n\) converges by direct comparison. 4. Adding finitely many initial terms preserves convergence, so \(\sum a_n\) converges.

Since \(a_n/b_n\to0\), eventually \(0<a_n<b_n\). The convergent upper comparison \(\sum b_n\) proves that \(\sum a_n\) converges.

- Check denominator signs before cross-multiplying. - Solve the resulting polynomial inequality on the relevant domain. - Test the preceding integer to establish minimality.

1. For \(n>3\), both denominators are positive. Cross-multiplying gives \(n^2\le2(n^2-3n)\). 2. This simplifies to \(n(n-6)\ge0\), so the inequality holds for \(n\ge6\). 3. It fails at \(n=5\), so the least integer is \(N=6\). 4. The tail is bounded above by \(2/n^2\), a convergent series. Therefore the original series converges.

The least integer is \(N=6\), and the series converges.

- Use the upper coefficient bound for a convergence argument. - Use the lower coefficient bound for a divergence argument. - Compare both cases with the same reciprocal-power family.

1. If \(p>1\), then \(0<\frac{c_n}{n^p}\le\frac{7}{n^p}\), and the upper comparison series converges. 2. If \(p\le1\), then \(\frac{c_n}{n^p}\ge\frac{2}{n^p}\), and the lower comparison series diverges. 3. Therefore the series converges exactly when \(p>1\).

The series converges exactly when \(p>1\), and diverges when \(p\le1\).

- Try the claimed inequality for both odd and even indices greater than one. - Check the sign requirement behind the comparison theorem. - Apply separate tests to the signed series and its absolute values.

1. The stated inequality itself is false for even \(n>1\), since \(1/n>1/n^2\). It holds for odd \(n\) only because the left side is negative. 2. In addition, direct comparison in this form requires nonnegative terms; it cannot use alternating terms in the proposed way. 3. The magnitudes \(1/n\) decrease to zero, so the signed series converges by the alternating series test. 4. The absolute-value series is harmonic and diverges, so the original series converges conditionally.

The inequality is false at even indices \(n>1\), and direct comparison cannot be applied to signed terms as proposed. The series converges conditionally.

- Replace the varying ratio bound by a fixed number below one on a tail. - Iterate the fixed bound to compare term magnitudes with a geometric sequence. - Add the finite initial terms after controlling the tail.

1. For \(n\ge2\), \(1/(n+2)\le1/4\), so \(|a_{n+1}/a_n|\le3/4\). 2. Repeatedly applying the bound gives \(|a_{2+k}|\le|a_2|(3/4)^k\). 3. The magnitude tail is bounded by a convergent geometric series. 4. Therefore \(\sum|a_n|\) converges, so \(\sum a_n\) converges absolutely.

The series converges absolutely.

- Recall exactly what a zero term limit does and does not prove about a series. - Use the dominant powers of \(n\) to choose a positive benchmark series. - For limit comparison, form the quotient of the given term and the benchmark term and simplify before taking the limit.

1. A term limit of \(0\) is necessary for convergence but does not prove convergence. 2. Compare with \(b_n=1/n^2\). Then \(\lim_{n\to\infty}\frac{(3n+2)/(n^3+1)}{1/n^2}=\lim_{n\to\infty}\frac{3n^3+2n^2}{n^3+1}=3\). 3. The limit is positive and finite, and \(\sum 1/n^2\) converges. 4. Therefore the given series converges by limit comparison.

The error is treating a zero term limit as sufficient for convergence. The series converges by limit comparison with \(\sum 1/n^2\), with comparison limit \(3\).

- Which familiar positive series has terms of a similar size? - Can you find a useful inequality or a positive finite comparison limit? - Check that the test you choose supports the conclusion you draw.

1. The harmonic series \(\sum1/n\) is a useful divergent benchmark. 2. Direct comparison works: \(\frac{3n+1}{n(n+1)}\ge\frac1n\) for \(n\ge1\), because \(3n+1\ge n+1\). 3. Alternatively, the ratio to \(1/n\) is \(\frac{3n+1}{n+1}\to3>0\), so limit comparison also works. 4. Either valid comparison proves the given series diverges.

The series diverges. For example, \(\frac{3n+1}{n(n+1)}\ge\frac1n\), and the harmonic series diverges. Limit comparison with ratio limit \(3\) also works.

- Use the difference of the dominant exponents to choose a benchmark of the form \(n^{p-q}\). - Compute the ratio after factoring \(n^p\) from the numerator and \(n^q\) from the denominator. - Check the boundary cases \(p=0\) or \(q=0\); the ratio need only approach a positive finite constant, not necessarily \(1\).

1. Compare with \(b_n=n^{p-q}=\frac1{n^{q-p}}\). 2. The ratio is \(\frac{a_n}{b_n}=\frac{1+n^{-p}}{1+2n^{-q}}\). 3. Because \(p,q\ge0\), the numerator tends to either \(1\) or \(2\), and the denominator tends to either \(1\) or \(3\). Thus the ratio limit is always positive and finite. 4. The benchmark \(\sum1/n^{q-p}\) converges exactly when \(q-p>1\). Therefore the given series converges when \(q-p>1\) and diverges when \(q-p\le1\).

Use \(b_n=n^{p-q}=1/n^{q-p}\). The ratio \(a_n/b_n\) approaches a positive finite constant, so the given series has the same behavior as \(\sum_{n=1}^{\infty}1/n^{q-p}\). It converges when \(q-p>1\) and diverges when \(q-p\le1\).

- Rearrange the inequality to isolate the required lower bound on \(C\). - The least valid constant is the maximum of \(\frac{n^2}{n^2-4}\) over integers \(n\ge3\). - Show that the corresponding real-variable function decreases for \(x>2\), then evaluate it at the first allowed integer.

1. Because all denominators are positive for \(n\ge3\), the inequality is equivalent to \(C\ge\frac{n^2}{n^2-4}\). 2. For \(g(x)=\frac{x^2}{x^2-4}\), \(g'(x)=-\frac{8x}{(x^2-4)^2}<0\) for \(x>2\). 3. Thus the largest required value occurs at the least allowed integer, \(n=3\). 4. Therefore the least constant is \(C=\frac{3^2}{3^2-4}=\frac95\).

\(C=\frac95\).

- Use the upper inequality in one direction. - Rearrange the lower inequality to obtain an upper comparison in the reverse direction. - Finite initial terms do not affect the conclusion.

1. If \(\sum b_n\) converges, then \(a_n\le Cb_n\) on a tail, so \(\sum a_n\) converges by direct comparison. 2. If \(\sum a_n\) converges, then \(b_n\le a_n/c\) on a tail, so \(\sum b_n\) converges. 3. Therefore convergence of either series implies convergence of the other. 4. Equivalently, divergence of either implies divergence of the other.

The bounds \(a_n\le Cb_n\) and \(b_n\le a_n/c\) transfer convergence in both directions, so the two positive series have the same convergence behavior.

- Turn closeness to a positive finite limit into two fixed inequalities. - Choose bounds that remain positive on a tail. - Use the two inequalities to transfer convergence behavior in both directions.

1. Choose \(N\) so that \(|a_n/b_n-L|<L/2\) for every \(n\ge N\). 2. Then \(L/2<a_n/b_n<3L/2\), so \((L/2)b_n<a_n<(3L/2)b_n\) on the tail. 3. These two constant-multiple bounds allow direct comparison in both directions. 4. Hence the two series either both converge or both diverge.

For sufficiently large \(n\), \(\frac L2 b_n<a_n<\frac{3L}{2}b_n\). These fixed positive bounds transfer convergence in both directions, so the series share the same behavior.

- Identify which denominator power dominates for large indices. - Treat the equal-exponent case separately when computing the constant. - Classify the benchmark using the larger exponent.

1. Let \(r=\max(p,q)\). The denominator is governed by \(n^r\). 2. If \(p\ne q\), then \(\frac{1/(n^p+n^q)}{1/n^r}\to1\). If \(p=q=r\), the ratio is \(1/2\). 3. In every case, the limiting ratio with \(1/n^r\) is positive and finite. 4. Therefore the series converges exactly when \(r>1\), or \(\max(p,q)>1\).

With \(r=\max(p,q)\), comparison to \(1/n^r\) has limit \(1\) when \(p\ne q\), or \(1/2\) when \(p=q\). Thus the series converges exactly when \(\max(p,q)>1\), and diverges otherwise.

- Fix one familiar convergent benchmark. - Choose one larger reciprocal power series that still converges. - Choose another larger reciprocal power series that crosses the divergence boundary.

1. Choose \(b_n=1/n^2\), so \(\sum b_n\) converges. 2. If \(a_n=1/n^{3/2}\), then \(a_n/b_n=\sqrt n\to\infty\), but \(\sum a_n\) converges. 3. If \(a_n=1/n\), then \(a_n/b_n=n\to\infty\), but \(\sum a_n\) diverges. 4. Therefore an infinite ratio against a convergent benchmark gives no conclusion by itself.

No. With \(b_n=1/n^2\), the choice \(a_n=1/n^{3/2}\) gives convergence, while \(a_n=1/n\) gives divergence; both ratios tend to infinity.

- Choose one sequence much smaller than the benchmark whose total is finite. - Choose another sequence still smaller than the benchmark whose total is infinite. - What do the two examples show about a zero limiting ratio in this situation?

1. Choose \(a_n=1/n^2\). Then \(a_n/b_n=1/n\to0\), and \(\sum a_n\) converges. 2. Define \(c_1=1\). For \(n\ge2\), choose \(c_n=1/(n\ln n)\). Then \(c_n/b_n=1/\ln n\to0\). 3. The series \(\sum_{n=2}^{\infty}1/(n\ln n)\) diverges, for example by the integral test. 4. Thus a ratio of \(0\) relative to a divergent comparison series can occur with either convergence or divergence, so that information is inconclusive.

One valid pair is \(a_n=1/n^2\), \(c_1=1\), and \(c_n=1/(n\ln n)\) for \(n\ge2\). The shared ratio condition does not determine convergence when the benchmark diverges.