Advanced Math Solutions - Series Convergence Calculator, Series Ratio Test

In our Series blogs, we’ve gone over four types of series, Geometric, p, Alternating, and Telescoping, and their convergence tests. Now, we will focus on convergence tests for any type of infinite series, as long as they meet the tests’ criteria. We will start off with the Series Ratio Test.

What does the Series Ratio Test do?

The goal of the Series Ratio Test is to determine if the series converges or diverges by evaluating the ratio of the general term of the series to its following term. The test determines if the ratio absolutely converges. A series absolutely convergences if the sum of the absolute value of the terms is finite. If there is absolute convergence, then there is convergence. This will make more sense, once you see the test and try out a few examples.

Some caveats: The test will not determine what the series will converge to. The test may also result in inconclusive results.

What is the Series Ratio Test?

Given a series, ∑a_n , determine L such that

                                             lim_{n→∞} \mid\frac{a_{n+1}}{a_n}\mid=L,where a_n≠0
If L<1, then ∑a_n  converges.
If L>1, then ∑a_n  diverges.
If L=1, then the test is inconclusive.

The test may seem pretty straight forward and simple, but determining what type of series to use this test on is not.

What types of series should the test be used on?

While there is no straight forward answer to this question, this test is typically helpful when determining convergence for series with exponential functions or factorials.

Now that we know what the series ratio test is, let’s see some examples of how it is used.

First example (click here):

                                                             ∑_{n=0}^∞\frac{2^n}{n!}

1. Determine the limit of lim_{n→∞} \mid\frac{a_{n+1}}{a_n} \mid

                 lim_{n→∞} \mid\frac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}}\mid=lim_{n→∞} \mid\frac{2^{n+1}∙n!}{(n+1)!∙2^n }\mid
                                                   =lim_{n→∞} \mid\frac{2∙n!}{(n+1)!}\mid

                                                             (Note: 2^{n+1} = 2^n ∙ 2^1 )

                                                         =lim_{n→∞} \mid\frac{2}{(n+1) }\mid

                                                (Note: \frac{n!}{(n+m)!}=\frac{1}{(n+1)∙(n+2)  ∙ ∙ ∙(n+m) })

                                                                         =0=L

For detailed procedures on how to determine the limit, please see the step by step procedures in the link.

2. Given L, determine convergence

                                                 Since L = 0 and is < 1, the series converges.

Next example (click here):

                                                                  ∑_{n=1}^∞\frac{6^n}{n}

1. Determine the limit of lim_{n→∞} \mid\frac{a_{n+1}}{a_n} \mid

                  lim_{n→∞} \mid\frac{\frac{6^{n+1}}{n+1}}{\frac{6^n}{n}}\mid=lim_{n→∞} \mid\frac{6^{n+1}∙n}{(n+1)∙6^n }\mid

                                                        =lim_{n→∞} \mid\frac{6n}{(n+1) }\mid

                                                       =6∙lim_{n→∞} \mid\frac{1}{(1+\frac{1}{n}) }\mid

                                                            (Note: \frac{n}{n+1 }=  \frac{1}{1+\frac{1}{n}})

                                                                         =6∙1=6=L

2. Given L, determine convergence

                                                  Since L = 6 and is > 1, the series diverges.


Next example (click here):

                                                                 ∑_{n=0}^∞\frac{2^2n}{3^2n}

1. Determine the limit of lim_{n→∞} \mid\frac{a_{n+1}}{a_n} \mid

 lim_{n→∞} \mid\frac{\frac{2^{2(n+1)}}{3^{2(n+1)}} }{\frac{2^2n}{3^2n} }\mid=lim_{n→∞} \mid\frac{(\frac{2}{3})^{2(n+1)}}{(\frac{2}{3})^2n} \mid

                                                       =lim_{n→∞} \mid(\frac{2}{3})^{2(n+1)-2n} \mid

                                                               =lim_{n→∞} \mid(\frac{2}{3})^2 \mid

                                                                               =\frac{4}{9}=L

2. Given L, determine convergence

                                                    Since L = \frac{4}{9} and is < 1, the series converges.

Last example:
                                                                      ∑_{n=1}^∞\frac{1}{n}


1. Determine the limit of lim_{n→∞} \mid\frac{a_{n+1}}{a_n} \mid

                                      lim_{n→∞} \mid\frac{\frac{1}{n+1}}{\frac{1}{n}}\mid=lim_{n→∞} \mid\frac{n}{n+1}\mid


                                                             =lim_{n→∞} \mid\frac{1}{1+\frac{1}{n}}\mid

                                                (Click here for the limit step by step procedures)

                                                                                 =1=L


2. Given L, determine convergence

                                                         Since L = 1, the test is inconclusive.

You may be wondering what to do if the ratio series test is inconclusive. In the next couple of blog posts, we will be discussing other convergence tests that can be used when the ratio test is inconclusive. For more practice on the Ratio Series Test, check out Symbolab’s Practice.

Until next time,

Leah