5.6 Ratio and root tests (Page 3/8)

Calculus volume 2 Page 3 / 8

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Using convergence tests

For each of the following series, determine which convergence test is the best to use and explain why. Then determine if the series converges or diverges. If the series is an alternating series, determine whether it converges absolutely, converges conditionally, or diverges.

$\sum_{n = 1}^{\infty} \frac{n^{2} + 2 n}{n^{3} + 3 n^{2} + 1}$
$\sum_{n = 1}^{\infty} \frac{{(−1)}^{n + 1} (3 n + 1)}{n!}$
$\sum_{n = 1}^{\infty} \frac{e^{n}}{n^{3}}$
$\sum_{n = 1}^{\infty} \frac{3^{n}}{{(n + 1)}^{n}}$

Step 1. The series is not a $p - series$ or geometric series.
Step 2. The series is not alternating.
Step 3. For large values of $n,$ we approximate the series by the expression

$\frac{n^{2} + 2 n}{n^{3} + 3 n^{2} + 1} \approx \frac{n^{2}}{n^{3}} = \frac{1}{n} .$

Therefore, it seems reasonable to apply the comparison test or limit comparison test using the series $\sum_{n = 1}^{\infty} 1 / n .$ Using the limit comparison test, we see that

$lim_{n \to \infty} \frac{(n^{2} + 2 n) / (n^{3} + 3 n^{2} + 1)}{1 / n} = lim_{n \to \infty} \frac{n^{3} + 2 n^{2}}{n^{3} + 3 n^{2} + 1} = 1 .$

Since the series $\sum_{n = 1}^{\infty} 1 / n$ diverges, this series diverges as well.
Step 1.The series is not a familiar series.
Step 2. The series is alternating. Since we are interested in absolute convergence, consider the series

$\sum_{n = 1}^{\infty} \frac{3 n}{(n + 1)!} .$

Step 3. The series is not similar to a p -series or geometric series.
Step 4. Since each term contains a factorial, apply the ratio test. We see that

$lim_{n \to \infty} \frac{(3 (n + 1)) / (n + 1)!}{(3 n + 1) / n!} = lim_{n \to \infty} \frac{3 n + 3}{(n + 1)!} \cdot \frac{n!}{3 n + 1} = lim_{n \to \infty} \frac{3 n + 3}{(n + 1) (3 n + 1)} = 0 .$

Therefore, this series converges, and we conclude that the original series converges absolutely, and thus converges.
Step 1. The series is not a familiar series.
Step 2. It is not an alternating series.
Step 3. There is no obvious series with which to compare this series.
Step 4. There is no factorial. There is a power, but it is not an ideal situation for the root test.
Step 5. To apply the divergence test, we calculate that

$lim_{n \to \infty} \frac{e^{n}}{n^{3}} = \infty .$

Therefore, by the divergence test, the series diverges.
Step 1. This series is not a familiar series.
Step 2. It is not an alternating series.
Step 3. There is no obvious series with which to compare this series.
Step 4. Since each term is a power of $n,$ we can apply the root test. Since

$lim_{n \to \infty} \sqrt[n]{{(\frac{3}{n + 1})}^{n}} = lim_{n \to \infty} \frac{3}{n + 1} = 0,$

by the root test, we conclude that the series converges.

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For the series $\sum_{n = 1}^{\infty} \frac{2^{n}}{3^{n} + n},$ determine which convergence test is the best to use and explain why.

The comparison test because $2^{n} / (3^{n} + n) < 2^{n} / 3^{n}$ for all positive integers $n .$ The limit comparison test could also be used.

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In [link] , we summarize the convergence tests and when each can be applied. Note that while the comparison test, limit comparison test, and integral test require the series $\sum_{n = 1}^{\infty} a_{n}$ to have nonnegative terms, if $\sum_{n = 1}^{\infty} a_{n}$ has negative terms, these tests can be applied to $\sum_{n = 1}^{\infty} | a_{n} |$ to test for absolute convergence.

Summary of convergence tests
Series or Test	Conclusions	Comments
Divergence Test For any series $\sum_{n = 1}^{\infty} a_{n},$ evaluate $lim_{n \to \infty} a_{n} .$	If $lim_{n \to \infty} a_{n} = 0,$ the test is inconclusive.	This test cannot prove convergence of a series.
If $lim_{n \to \infty} a_{n} \neq 0,$ the series diverges.
Geometric Series $\sum_{n = 1}^{\infty} a r^{n - 1}$	If $\| r \| < 1,$ the series converges to $a / (1 - r) .$	Any geometric series can be reindexed to be written in the form $a + a r + a r^{2} + ⋯,$ where $a$ is the initial term and $r$ is the ratio.
If $\| r \| \geq 1,$ the series diverges.
p -Series $\sum_{n = 1}^{\infty} \frac{1}{n^{p}}$	If $p > 1,$ the series converges.	For $p = 1,$ we have the harmonic series $\sum_{n = 1}^{\infty} 1 / n .$
If $p \leq 1,$ the series diverges.
Comparison Test For $\sum_{n = 1}^{\infty} a_{n}$ with nonnegative terms, compare with a known series $\sum_{n = 1}^{\infty} b_{n} .$	If $a_{n} \leq b_{n}$ for all $n \geq N$ and $\sum_{n = 1}^{\infty} b_{n}$ converges, then $\sum_{n = 1}^{\infty} a_{n}$ converges.	Typically used for a series similar to a geometric or $p$ -series. It can sometimes be difficult to find an appropriate series.
If $a_{n} \geq b_{n}$ for all $n \geq N$ and $\sum_{n = 1}^{\infty} b_{n}$ diverges, then $\sum_{n = 1}^{\infty} a_{n}$ diverges.
Limit Comparison Test For $\sum_{n = 1}^{\infty} a_{n}$ with positive terms, compare with a series $\sum_{n = 1}^{\infty} b_{n}$ by evaluating $L = lim_{n \to \infty} \frac{a_{n}}{b_{n}} .$	If $L$ is a real number and $L \neq 0,$ then $\sum_{n = 1}^{\infty} a_{n}$ and $\sum_{n = 1}^{\infty} b_{n}$ both converge or both diverge.	Typically used for a series similar to a geometric or $p$ -series. Often easier to apply than the comparison test.
If $L = 0$ and $\sum_{n = 1}^{\infty} b_{n}$ converges, then $\sum_{n = 1}^{\infty} a_{n}$ converges.
If $L = \infty$ and $\sum_{n = 1}^{\infty} b_{n}$ diverges, then $\sum_{n = 1}^{\infty} a_{n}$ diverges.
Integral Test If there exists a positive, continuous, decreasing function $f$ such that $a_{n} = f (n)$ for all $n \geq N,$ evaluate $\int_{N}^{\infty} f (x) d x .$	$\int_{N}^{\infty} f (x) d x$ and $\sum_{n = 1}^{\infty} a_{n}$ both converge or both diverge.	Limited to those series for which the corresponding function $f$ can be easily integrated.
Alternating Series $\sum_{n = 1}^{\infty} {(−1)}^{n + 1} b_{n} or \sum_{n = 1}^{\infty} {(−1)}^{n} b_{n}$	If $b_{n + 1} \leq b_{n}$ for all $n \geq 1$ and $b_{n} \to 0,$ then the series converges.	Only applies to alternating series.
Ratio Test For any series $\sum_{n = 1}^{\infty} a_{n}$ with nonzero terms, let $ρ = lim_{n \to \infty} \| \frac{a_{n + 1}}{a_{n}} \| .$	If $0 \leq ρ < 1,$ the series converges absolutely.	Often used for series involving factorials or exponentials.
If $ρ > 1 or ρ = \infty,$ the series diverges.
If $ρ = 1,$ the test is inconclusive.
Root Test For any series $\sum_{n = 1}^{\infty} a_{n},$ let $ρ = lim_{n \to \infty} \sqrt[n]{\| a_{n} \|} .$	If $0 \leq ρ < 1,$ the series converges absolutely.	Often used for series where $\| a_{n} \| = b_{n}^{n} .$
If $ρ > 1 or ρ = \infty,$ the series diverges.
If $ρ = 1,$ the test is inconclusive.

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Source: OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2

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