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Comparison Tests

Use the comparison test to test a series for convergence.

Comparison Test

In the preceding two sections, we discussed two large classes of series: geometric series and p-series. We know exactly when these series converge and when they diverge. Here we show how to use the convergence or divergence of these series to prove convergence or divergence for other series, using a method called the comparison test.

For example, consider the series

\[\sum _{n=1}^{\infty }\frac{1}{{n}^{2}+1}.\]

This series looks similar to the convergent series

\[\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}.\]

Since the terms in each of the series are positive, the sequence of partial sums for each series is monotone increasing. Furthermore, since

\[0<\frac{1}{{n}^{2}+1}<\frac{1}{{n}^{2}}\]

for all positive integers \(n,\) the \(k\text{th}\) partial sum \({S}_{k}\) of \(\sum _{n=1}^{\infty }\frac{1}{{n}^{2}+1}\) satisfies

\[{S}_{k}=\sum _{n=1}^{k}\frac{1}{{n}^{2}+1}<\sum _{n=1}^{k}\frac{1}{{n}^{2}}<\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}.\]

(See (a) and .) Since the series on the right converges, the sequence \(\{{S}_{k}\}\) is bounded above. We conclude that \(\{{S}_{k}\}\) is a monotone increasing sequence that is bounded above. Therefore, by the Monotone Convergence Theorem, \(\{{S}_{k}\}\) converges, and thus

\[\sum _{n=1}^{\infty }\frac{1}{{n}^{2}+1}\]

converges.

\[\sum _{n=1}^{\infty }\frac{1}{n-1\text{/}2}.\]\[\sum _{n=1}^{\infty }\frac{1}{n}.\]\[\frac{1}{n-1\text{/}2}>\frac{1}{n}>0\]\[{S}_{k}=\sum _{n=1}^{k}\frac{1}{n-1\text{/}2}>\sum _{n=1}^{k}\frac{1}{n}.\]\[\sum _{n=1}^{\infty }\frac{1}{n-1\text{/}2}\]

\(k\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)	\(7\)	\(8\)
\(\sum _{n=1}^{k}\frac{1}{{n}^{2}+1}\)	\(0.5\)	\(0.7\)	\(0.8\)	\(0.8588\)	\(0.8973\)	\(0.9243\)	\(0.9443\)	\(0.9597\)
\(\sum _{n=1}^{k}\frac{1}{{n}^{2}}\)	\(1\)	\(\begin{array}{l}1.25\end{array}\)	\(1.3611\)	\(1.4236\)	\(1.4636\)	\(1.4914\)	\(1.5118\)	\(1.5274\)

\(k\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)	\(7\)	\(8\)
\(\sum _{n=1}^{k}\frac{1}{n-1\text{/}2}\)	\(2\)	\(2.6667\)	\(3.0667\)	\(3.3524\)	\(3.5746\)	\(3.7564\)	\(3.9103\)	\(4.0436\)
\(\sum _{n=1}^{k}\frac{1}{n}\)	\(1\)	\(1.5\)	\(\begin{array}{l}1.8333\end{array}\)	\(2.0933\)	\(2.2833\)	\(2.45\)	\(2.5929\)	\(2.7179\)

Condensed: the full section is in OpenStax Calculus Volume 2.

Limit Comparison Test

The comparison test works nicely if we can find a comparable series satisfying the hypothesis of the test. However, sometimes finding an appropriate series can be difficult. Consider the series

\[\sum _{n=2}^{\infty }\frac{1}{{n}^{2}-1}.\]

It is natural to compare this series with the convergent series

\[\sum _{n=2}^{\infty }\frac{1}{{n}^{2}}.\]

However, this series does not satisfy the hypothesis necessary to use the comparison test because

\[\frac{1}{{n}^{2}-1}>\frac{1}{{n}^{2}}\]

for all integers \(n\ge 2.\) Although we could look for a different series with which to compare \(\sum _{n=2}^{\infty }1\text{/}({n}^{2}-1),\) instead we show how we can use the limit comparison test to compare

\[\sum _{n=2}^{\infty }\frac{1}{{n}^{2}-1}\ \text{and}\ \sum _{n=2}^{\infty }\frac{1}{{n}^{2}}.\]

Let us examine the idea behind the limit comparison test. Consider two series \(\sum _{n=1}^{\infty }{a}_{n}\) and \(\sum _{n=1}^{\infty }{b}_{n}.\) with positive terms \({a}_{n}\ \text{and}\ {b}_{n}\) and evaluate

\[\underset{n\to \infty }{\text{lim}}\frac{{a}_{n}}{{b}_{n}}.\]

\[\underset{n\to \infty }{\text{lim}}\frac{{a}_{n}}{{b}_{n}}=L\ne 0,\]

then, for \(n\) sufficiently large, \({a}_{n}\approx L{b}_{n}.\) Therefore, either both series converge or both series diverge. For the series \(\sum _{n=2}^{\infty }1\text{/}({n}^{2}-1)\) and \(\sum _{n=2}^{\infty }1\text{/}{n}^{2},\) we see that

\[\underset{n\to \infty }{\text{lim}}\frac{1\text{/}({n}^{2}-1)}{1\text{/}{n}^{2}}=\underset{n\to \infty }{\text{lim}}\ \frac{{n}^{2}}{{n}^{2}-1}=1.\]\[\sum _{n=2}^{\infty }\frac{1}{{n}^{2}-1}\]\[\underset{n\to \infty }{\text{lim}}\frac{{a}_{n}}{{b}_{n}}=0.\]\[\underset{n\to \infty }{\text{lim}}\frac{{a}_{n}}{{b}_{n}}=\infty .\]\[\frac{1\text{/}\sqrt{n}}{1\text{/}{n}^{3}}=\frac{{n}^{3}}{\sqrt{n}}={n}^{5\text{/}2}\to \infty \ \text{as}\ n\to \infty .\]\[\frac{1\text{/}{n}^{2}}{1\text{/}{n}^{3}}=n\to \infty \ \text{as}\ n\to \infty .\]

Condensed: the full section is in OpenStax Calculus Volume 2.

Key Concepts

The comparison tests are used to determine convergence or divergence of series with positive terms.
When using the comparison tests, a series \(\sum _{n=1}^{\infty }{a}_{n}\) is often compared to a geometric or p-series.

Comparison Tests

Use the comparison test to determine whether the following series converge.

Use the limit comparison test to determine whether each of the following series converges or diverges.

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Вежбај (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

For each of the following series, use the comparison test to determine whether the series converges or diverges.
1. \(\sum _{n=1}^{\infty }\frac{1}{{n}^{3}+3n+1}\)
2. \(\sum _{n=1}^{\infty }\frac{1}{{2}^{n}+1}\)
3. \(\sum _{n=2}^{\infty }\frac{1}{\text{ln}(n)}\)
Одкриј го одговорот
1. Compare to \(\sum _{n=1}^{\infty }\frac{1}{{n}^{3}}\) Since \(\sum _{n=1}^{\infty }\frac{1}{{n}^{3}}\) is a p-series with \(p=3,\) it converges. Further,
  \[\frac{1}{{n}^{3}+3n+1}<\frac{1}{{n}^{3}}\]
  for every positive integer \(n.\) Therefore, we can conclude that \(\sum _{n=1}^{\infty }\frac{1}{{n}^{3}+3n+1}\) converges.
2. Compare to \(\sum _{n=1}^{\infty }{(\frac{1}{2})}^{n}.\) Since \(\sum _{n=1}^{\infty }{(\frac{1}{2})}^{n}\) is a geometric series with \(r=1\text{/}2\) and \(|1\text{/}2|<1,\) it converges. Also,
  \[\frac{1}{{2}^{n}+1}<\frac{1}{{2}^{n}}\]
  for every positive integer \(n.\) Therefore, we see that \(\sum _{n=1}^{\infty }\frac{1}{{2}^{n}+1}\) converges.
3. Compare to \(\sum _{n=2}^{\infty }\frac{1}{n}.\) Since
  \[\frac{1}{\text{ln}(n)}>\frac{1}{n}\]
  for every integer \(n\ge 2\) and \(\sum _{n=2}^{\infty }1\text{/}n\) diverges, we have that \(\sum _{n=2}^{\infty }\frac{1}{\text{ln}(n)}\) diverges.
Use the comparison test to determine if the series \(\sum _{n=1}^{\infty }\frac{n}{{n}^{3}+n+1}\) converges or diverges.

Одкриј го одговорот

The series converges.
For each of the following series, use the limit comparison test to determine whether the series converges or diverges. If the test does not apply, say so.
1. \(\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}+1}\)
2. \(\sum _{n=1}^{\infty }\frac{{2}^{n}+1}{{3}^{n}}\)
3. \(\sum _{n=1}^{\infty }\frac{\text{ln}(n)}{{n}^{2}}\)
Одкриј го одговорот
1. Compare this series to \(\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}.\) Calculate
  \[\underset{n\to \infty }{\text{lim}}\ \frac{1\text{/}(\sqrt{n}+1)}{1\text{/}\sqrt{n}}=\underset{n\to \infty }{\text{lim}}\begin{array}{l}\frac{\sqrt{n}}{\sqrt{n}+1} \\ \end{array}=\underset{n\to \infty }{\text{lim}}\ \frac{1}{1+1\text{/}\sqrt{n}}=1.\] By the limit comparison test, since \(\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}\) diverges, then \(\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}+1}\) diverges.
2. Compare this series to \(\sum _{n=1}^{\infty }{(\frac{2}{3})}^{n}.\) We see that
  \[\underset{n\to \infty }{\text{lim}}\frac{({2}^{n}+1)\text{/}{3}^{n}}{{2}^{n}\text{/}{3}^{n}}=\underset{n\to \infty }{\text{lim}}\frac{{2}^{n}+1}{{3}^{n}}\cdot \frac{{3}^{n}}{{2}^{n}}=\underset{n\to \infty }{\text{lim}}\frac{{2}^{n}+1}{{2}^{n}}=\underset{n\to \infty }{\text{lim}}[1+{(\frac{1}{2})}^{n}]=1.\]
  Therefore,
  \[\underset{n\to \infty }{\text{lim}}\frac{({2}^{n}+1)\text{/}{3}^{n}}{{2}^{n}\text{/}{3}^{n}}=1.\]
  Since \(\sum _{n=1}^{\infty }{(\frac{2}{3})}^{n}\) converges, we conclude that \(\sum _{n=1}^{\infty }\frac{{2}^{n}+1}{{3}^{n}}\) converges.
3. Since \text{ln}\ n \[\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\ n\text{/}{n}^{2}}{1\text{/}n}=\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\ n}{{n}^{2}}\cdot \frac{n}{1}=\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\ n}{n}.\]
  In order to evaluate \(\underset{n\to \infty }{\text{lim}}\text{ln}\ n\text{/}n,\) evaluate the limit as \(x\to \infty\) of the real-valued function \(\text{ln}(x)\text{/}x.\) These two limits are equal, and making this change allows us to use L’Hôpital’s rule. We obtain
  \[\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\ x}{x}=\underset{x\to \infty }{\text{lim}}\frac{1}{x}=0.\]
  Therefore, \(\underset{n\to \infty }{\text{lim}}\text{ln}\ n\text{/}n=0,\) and, consequently,
  \[\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\ n\text{/}{n}^{2}}{1\text{/}n}=0.\]
  Since the limit is \(0\) but \(\sum _{n=1}^{\infty }\frac{1}{n}\) diverges, the limit comparison test does not provide any information.
  Compare with \(\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}\) instead. In this case,
  \[\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\ n\text{/}{n}^{2}}{1\text{/}{n}^{2}}=\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\ n}{{n}^{2}}\cdot \frac{{n}^{2}}{1}=\underset{n\to \infty }{\text{lim}}\text{ln}\ n=\infty .\]
  Since the limit is \(\infty\) but \(\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}\) converges, the test still does not provide any information.
  So now we try a series between the two we already tried. Choosing the series \(\sum _{n=1}^{\infty }\frac{1}{{n}^{3\text{/}2}},\) we see that
  \[\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\ n\text{/}{n}^{2}}{1\text{/}{n}^{3\text{/}2}}=\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\ n}{{n}^{2}}\cdot \frac{{n}^{3\text{/}2}}{1}=\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\ n}{\sqrt{n}}.\]
  As above, in order to evaluate \(\underset{n\to \infty }{\text{lim}}\text{ln}\ n\text{/}\sqrt{n},\) evaluate the limit as \(x\to \infty\) of the real-valued function \(\text{ln}\ x\text{/}\sqrt{x}.\) Using L’Hôpital’s rule,
  \[\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\ x}{\sqrt{x}}=\underset{x\to \infty }{\text{lim}}\frac{2\sqrt{x}}{x}=\underset{x\to \infty }{\text{lim}}\frac{2}{\sqrt{x}}=0.\]
  Since the limit is \(0\) and \(\sum _{n=1}^{\infty }\frac{1}{{n}^{3\text{/}2}}\) converges, we can conclude that \(\sum _{n=1}^{\infty }\frac{\text{ln}\ n}{{n}^{2}}\) converges.
Use the limit comparison test to determine whether the series \(\sum _{n=1}^{\infty }\frac{{5}^{n}}{{3}^{n}+2}\) converges or diverges.

Одкриј го одговорот

The series diverges.
\(\sum _{n=1}^{\infty }{a}_{n}\) where \({a}_{n}=\frac{2}{n(n+1)}\)
\(\sum _{n=1}^{\infty }{a}_{n}\) where \({a}_{n}=\frac{1}{n(n+1\text{/}2)}\)

Одкриј го одговорот

Converges by comparison with \(1\text{/}{n}^{2}.\)
\(\sum _{n=1}^{\infty }\frac{1}{2(n+1)}\)
\(\sum _{n=2}^{\infty }\frac{1}{2\left(n-1\right)}\)

Одкриј го одговорот

The sum is one half times a harmonic series, which diverges.
\(\sum _{n=2}^{\infty }\frac{1}{{(n\ \text{ln}\ n)}^{2}}\)
\(\sum _{n=1}^{\infty }\frac{n\text{!}}{(n+2)\text{!}}\)

Одкриј го одговорот

\({a}_{n}=1\text{/}(n+1)(n+2)<1\text{/}{n}^{2}.\) Converges by comparison with p-series, \(p=2.\)
\(\sum _{n=1}^{\infty }\frac{1}{n\text{!}}\)
\(\sum _{n=1}^{\infty }\frac{\text{sin}(1\text{/}n)}{{n}^{2}}\)

Одкриј го одговорот

\(\text{sin}(1\text{/}n)\le 1\text{/}n,\) so converges by comparison with p-series, \(p=2.\)
\(\sum _{n=1}^{\infty }\frac{{\text{sin}}^{2}n}{{n}^{2}}\)
\(\sum _{n=1}^{\infty }\frac{\text{sin}(1\text{/}n)}{(\sqrt{n}{)}^{3}}\)

Одкриј го одговорот

\(\text{sin}(1\text{/}n)\le 1,\) so converges by comparison with p-series, \(p=3\text{/}2.\)
\(\sum _{n=1}^{\infty }\frac{{n}^{1.2}-1}{{n}^{2.3}+1}\)
\(\sum _{n=1}^{\infty }\frac{\sqrt{n+1}-\sqrt{n}}{n}\)

Одкриј го одговорот

Since \(\sqrt{n+1}-\sqrt{n}=1\text{/}(\sqrt{n+1/2}+\sqrt{n})\le 2\text{/}\sqrt{n},\) series converges by comparison with p-series for \(p=1.5.\)
\(\sum _{n=1}^{\infty }\frac{\sqrt[4]{n}}{\sqrt[3]{{n}^{4}+{n}^{2}}}\)
\(\sum _{n=1}^{\infty }{(\frac{\text{ln}\ n}{n})}^{2}\)

Одкриј го одговорот

Converges by limit comparison with p-series for \(p>1.\)
\(\sum _{n=1}^{\infty }{(\frac{\text{ln}\ n}{{n}^{0.6}})}^{2}\)
\(\sum _{n=1}^{\infty }\frac{\text{ln}(1+\frac{1}{n})}{n}\)

Одкриј го одговорот

Converges by limit comparison with p-series, \(p=2.\)
\(\sum _{n=1}^{\infty }\text{ln}(1+\frac{1}{{n}^{2}})\)
\(\sum _{n=1}^{\infty }\frac{1}{{4}^{n}-{3}^{n}}\)

Одкриј го одговорот

Converges by limit comparison with \({4}^{\text{-}n}.\)
\(\sum _{n=1}^{\infty }\frac{1}{{n}^{2}-n\ \text{sin}\ n}\)
\(\sum _{n=1}^{\infty }\frac{1}{{e}^{(1.1)n}-{3}^{n}}\)

Одкриј го одговорот

Converges by limit comparison with \(1\text{/}{e}^{1.1n}.\)
\(\sum _{n=1}^{\infty }\frac{1}{{e}^{(1.01)n}-{3}^{n}}\)
\(\sum _{n=1}^{\infty }\frac{1}{{n}^{1+1\text{/}n}}\)

Одкриј го одговорот

Diverges by limit comparison with harmonic series.
\(\sum _{n=1}^{\infty }\frac{1}{{2}^{1+1\text{/}n}{n}^{1+1\text{/}n}}\)
\(\sum _{n=1}^{\infty }(\frac{1}{n}-\text{sin}(\frac{1}{n}))\)

Одкриј го одговорот

Converges by limit comparison with p-series, \(p=3.\)
\(\sum _{n=1}^{\infty }(1-\text{cos}(\frac{1}{n}))\)
\(\sum _{n=1}^{\infty }\frac{1}{n}(\frac{\pi }{2}-{\text{tan}}^{-1}n)\)

Одкриј го одговорот

Converges by limit comparison with p-series, \(p=2.\)
\(\sum _{n=1}^{\infty }{(1-\frac{1}{n})}^{n}\) (Hint:\({(1-\frac{1}{n})}^{n}\to 1\text{/}e.)\)
\(\sum _{n=1}^{\infty }(1-{e}^{-1\text{/}n})\) (Hint:\(1\text{/}e\approx {(1-1\text{/}n)}^{n},\) so \(1-{e}^{-1\text{/}n}\approx 1\text{/}n.)\)

Одкриј го одговорот

Diverges by limit comparison with \(1\text{/}n.\)
Does \(\sum _{n=2}^{\infty }\frac{1}{{(\text{ln}\ n)}^{p}}\) converge if \(p\) is large enough? If so, for which \(p\text{?}\)
Does \(\sum _{n=1}^{\infty }{(\frac{(\text{ln}\ n)}{n})}^{p}\) converge if \(p\) is large enough? If so, for which \(p\text{?}\)

Одкриј го одговорот

Converges for \(p>1\) by comparison with a \(p\) series for slightly smaller \(p.\)
For which \(p\) does the series \(\sum _{n=1}^{\infty }{2}^{pn}\text{/}{3}^{n}\) converge?
For which \(p>0\) does the series \(\sum _{n=1}^{\infty }\frac{{n}^{p}}{{2}^{n}}\) converge?

Одкриј го одговорот

Converges for all \(p>0.\)
For which \(r>0\) does the series \(\sum _{n=1}^{\infty }\frac{{r}^{{n}^{2}}}{{2}^{n}}\) converge?
For which \(r>0\) does the series \(\sum _{n=1}^{\infty }\frac{{2}^{n}}{{r}^{{n}^{2}}}\) converge?

Одкриј го одговорот

Converges for all \(r>1.\) If \(r>1\) then \({r}^{n}>4,\) say, once \(n>\text{ln}(2)\text{/}\text{ln}(r)\) and then the series converges by limit comparison with a geometric series with ratio \(1\text{/}2.\)
Find all values of \(p\) and \(q\) such that \(\sum _{n=1}^{\infty }\frac{{n}^{p}}{{(n\text{!})}^{q}}\) converges.
Does \(_{n=1}^{\infty }\left(\frac{n\pi }{2}\right)\) converge or diverge? Explain.

Одкриј го одговорот

The numerator is equal to \(1\) when \(n\) is odd and \(0\) when \(n\) is even, so the series can be rewritten \(\sum _{n=1}^{\infty }\frac{1}{2n+1},\) which diverges by limit comparison with the harmonic series.

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Како да го направиш тоа?: Comparison Tests

Use the comparison test to test a series for convergence.
Use the limit comparison test to determine convergence of a series.
Suppose there exists an integer
Suppose there exists an integer
Compare to
Compare to
Compare to
If

Прашања што луѓето ги поставуваат

What is a derivative in one sentence?

The slope of the graph at a point, the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate: the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function, a polynomial times an exponential, log or trig function.

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