Ratio and Root Tests: Learn It 3

Giselle Miller

Ratio and Root Tests: Learn It 3

Choosing a Convergence Test

You now have several convergence tests in your toolkit, but no single test works for every series. The key is developing a systematic strategy to choose the most effective test for each situation.

Problem-Solving Strategy: Choosing a Convergence Test for a Series

When analyzing a series [latex]\displaystyle\sum_{n=1}^{\infty}{a}_{n}[/latex], work through these steps:

Step 1: Check for familiar series Is this a series you recognize?

Harmonic series [latex]\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}[/latex] (diverges)
Alternating harmonic series [latex]\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}[/latex] (converges)
Geometric series [latex]\displaystyle\sum_{n=1}^{\infty}ar^{n-1}[/latex] (check if [latex]|r|<1[/latex])
p-series [latex]\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^p}[/latex] (check if [latex]p>1[/latex])

Step 2: Check for alternating series Does the series have terms that alternate in sign?

If you only need to know whether it converges, use the alternating series test
If you need absolute convergence, proceed to Step 3 using [latex]\displaystyle\sum_{n=1}^{\infty}|a_{n}|[/latex]

Step 3: Look for comparison opportunities Does the series behave similarly to a p-series or geometric series?

Try the comparison test or limit comparison test

Step 4: Check for factorials or exponentials Do the terms contain factorials ([latex]n![/latex]) or exponential expressions?

If [latex]a_{n} = b_{n}^{n}[/latex], try the root test first
Otherwise, try the ratio test first

Step 5: Use remaining tests

Apply the divergence test (if [latex]\lim_{n\to\infty}a_{n} \neq 0[/latex], the series diverges)
If the divergence test is inconclusive, try the integral test

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.

[latex]\displaystyle\sum _{n=1}^{\infty }\frac{{n}^{2}+2n}{{n}^{3}+3{n}^{2}+1}[/latex]
[latex]\displaystyle\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}\left(3n+1\right)}{n\text{!}}[/latex]
[latex]\displaystyle\sum _{n=1}^{\infty }\frac{{e}^{n}}{{n}^{3}}[/latex]
[latex]\displaystyle\sum _{n=1}^{\infty }\frac{{3}^{n}}{{\left(n+1\right)}^{n}}[/latex]

Step 1. The series is not a [latex]p-\text{series}[/latex] or geometric series.

Step 2. The series is not alternating.

Step 3. For large values of [latex]n[/latex], we approximate the series by the expression

[latex]\frac{{n}^{2}+2n}{{n}^{3}+3{n}^{2}+1}\approx \frac{{n}^{2}}{{n}^{3}}=\frac{1}{n}[/latex].

Therefore, it seems reasonable to apply the comparison test or limit comparison test using the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}[/latex]. Using the limit comparison test, we see that

[latex]\underset{n\to \infty }{\text{lim}}\frac{\frac{\left({n}^{2}+2n\right)}{\left({n}^{3}+3{n}^{2}+1\right)}}{\frac{1}{n}}=\underset{n\to \infty }{\text{lim}}\frac{{n}^{3}+2{n}^{2}}{{n}^{3}+3{n}^{2}+1}=1[/latex].

Since the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}[/latex] 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

[latex]\displaystyle\sum _{n=1}^{\infty }\frac{3n}{\left(n+1\right)\text{!}}[/latex].

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

[latex]\underset{n\to \infty }{\text{lim}}\frac{\frac{\left(3\left(n+1\right)\right)}{\left(n+1\right)\text{!}}}{\frac{\left(3n+1\right)}{n\text{!}}}=\underset{n\to \infty }{\text{lim}}\frac{3n+3}{\left(n+1\right)\text{!}}\cdot \frac{n\text{!}}{3n+1}=\underset{n\to \infty }{\text{lim}}\frac{3n+3}{\left(n+1\right)\left(3n+1\right)}=0[/latex].

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

[latex]\underset{n\to \infty }{\text{lim}}\frac{{e}^{n}}{{n}^{3}}=\infty[/latex].

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 [latex]n[/latex], we can apply the root test. Since

[latex]\underset{n\to \infty }{\text{lim}}\sqrt[n]{{\left(\frac{3}{n+1}\right)}^{n}}=\underset{n\to \infty }{\text{lim}}\frac{3}{n+1}=0[/latex],

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

Summary of Convergence Tests

The table below summarizes all convergence tests and their applications. Remember that the comparison test, limit comparison test, and integral test require nonnegative terms. If your series has negative terms, apply these tests to [latex]\displaystyle\sum_{n=1}^{\infty}|a_{n}|[/latex] to check for absolute convergence.

Test	When to Use	Conclusions	Key Notes
Divergence Test	Any series [latex]\displaystyle\sum_{n=1}^{\infty}a_{n}[/latex]	If [latex]\lim_{n\to \infty}a_{n} \neq 0[/latex], series diverges If [latex]\lim_{n\to \infty}a_{n} = 0[/latex], test is inconclusive	Cannot prove convergence
Geometric Series	[latex]\displaystyle\sum_{n=1}^{\infty}ar^{n-1}[/latex]	If [latex]\|r\| < 1[/latex], converges to [latex]\frac{a}{1-r}[/latex] If [latex]\|r\| \geq 1[/latex], diverges	Look for constant ratio between terms
p-Series	[latex]\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^p}[/latex]	If [latex]p > 1[/latex], converges If [latex]p \leq 1[/latex], diverges	[latex]p = 1[/latex] gives harmonic series
Comparison Test	Nonnegative terms, similar to known series	If [latex]a_n \leq b_n[/latex] and [latex]\sum b_n[/latex] converges, then [latex]\sum a_n[/latex] converges If [latex]a_n \geq b_n[/latex] and [latex]\sum b_n[/latex] diverges, then [latex]\sum a_n[/latex] diverges	Need to find good comparison series
Limit Comparison Test	Positive terms, similar to known series	If [latex]L = \lim_{n\to\infty}\frac{a_n}{b_n}[/latex] exists and [latex]L \neq 0[/latex], both series have same behavior Special cases: [latex]L = 0[/latex] or [latex]L = \infty[/latex]	Often easier than comparison test
Integral Test	Positive, continuous, decreasing function [latex]f[/latex] where [latex]a_n = f(n)[/latex]	[latex]\int_N^{\infty}f(x)dx[/latex] and [latex]\sum a_n[/latex] both converge or both diverge	Limited to easily integrable functions
Alternating Series Test	[latex]\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}b_n[/latex] or [latex]\displaystyle\sum_{n=1}^{\infty}(-1)^n b_n[/latex]	If [latex]b_{n+1} \leq b_n[/latex] and [latex]b_n \to 0[/latex], series converges	Only for alternating series
Ratio Test	Series with factorials or exponentials	Let [latex]\rho = \lim_{n\to\infty}\left\|\frac{a_{n+1}}{a_n}\right\|[/latex] If [latex]\rho < 1[/latex], converges absolutely If [latex]\rho > 1[/latex], diverges If [latex]\rho = 1[/latex], inconclusive	Great for factorials
Root Test	Series where [latex]\|a_n\| = b_n^n[/latex]	Let [latex]\rho = \lim_{n\to\infty}\sqrt[n]{\|a_n\|}[/latex] If [latex]\rho < 1[/latex], converges absolutely If [latex]\rho > 1[/latex], diverges If [latex]\rho = 1[/latex], inconclusive	Best for [latex]n[/latex]th powers