Ratio Test for Convergence (College Board AP® Calculus BC): Study Guide

Use the ratio test for series whose terms include more complicated expressions like exponentials, logarithms and factorials. The ratio test will generally be inconclusive for series with terms that are rational functions with only polynomials in the numerator and/or denominator.

Apply the ratio test to each of the following series.

(a) $\sum _{n=1}^{\infty }\frac{n!}{e^n}$

(b)$\sum _{n=1}^{\infty }{\left(-1\right)}^n\frac{{\left(\ln 2\right)}^n}{n^4}$

(c) $\sum _{n=1}^{\infty }\frac{1}{n^2}$

Answer:

(a)

All the terms here are positive, so we won't need the absolute value sign when taking the limit

Write and simplify the ratio $\frac{a_{n+1}}{a_n}$

$\begin{array}{rcl}\frac{\frac{\left(n+1\right)!}{e^{n+1}}}{\frac{n!}{e^n}}& =& \frac{\left(n+1\right)!}{e^{n+1}}·\frac{e^n}{n!}\\ & =& \frac{\left(n+1\right)·\cancel{n!}·\cancel{e^n}}{e·\cancel{e^n}·\cancel{n!}}\\ & =& \frac{n+1}{e}\end{array}$

Take the limit; remember that $e=2.718281...$ is just a constant

$\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=\underset{n\to \infty }{\lim }\frac{n+1}{e}=\infty >1$

The limit is greater than 1 (it diverges to positive infinity), so the series is divergent

According to the ratio test the series diverges

(b)

Write and simplify the ratio $\frac{a_{n+1}}{a_n}$

$\begin{array}{rcl}\frac{{\left(-1\right)}^{n+1}\frac{{\left(\ln 2\right)}^{n+1}}{{\left(n+1\right)}^4}}{{\left(-1\right)}^n\frac{{\left(\ln 2\right)}^n}{n^4}}& =& \frac{{\left(-1\right)}^{n+1}·{\left(\ln 2\right)}^{n+1}}{{\left(n+1\right)}^4}·\frac{n^4}{{\left(-1\right)}^n·{\left(\ln 2\right)}^n}\\ & =& \frac{{\left(-1\right)}^{n+1}}{{\left(-1\right)}^n}·\frac{\ln 2·\cancel{{\left(\ln 2\right)}^n}·n^4}{\cancel{{\left(\ln 2\right)}^n}·{\left(n+1\right)}^4}\\ & =& \left(-1\right)·\ln 2·{\left(\frac{n}{n+1}\right)}^4\\ & =& -\ln 2·{\left(\frac{n}{n+1}\right)}^4\\ & =& -\ln 2·{\left(\frac{n}{n+1}·\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}}\right)}^4\\ & =& -\ln 2·{\left(\frac{1}{1+{\displaystyle \frac{1}{n}}}\right)}^4\end{array}$

Take the limit

$\begin{array}{rcl}\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|& =& \underset{n\to \infty }{\lim }\left|-\ln 2·{\left(\frac{1}{1+{\displaystyle \frac{1}{n}}}\right)}^4\right|\\ & =& \underset{n\to \infty }{\lim }\ln 2·{\left(\frac{1}{1+{\displaystyle \frac{1}{n}}}\right)}^4\\ & =& \ln 2·{\left(\frac{1}{1+0}\right)}^4\\ & =& \ln 2\\ & =& 0.693147...<1\end{array}$

The limit is less than 1, so the series converges

Note that if $\ln 2$ in the series term formula were changed to $\ln 3=1.098612...$, then the series would diverge!

According to the ratio test the series converges

(c)

This is a p-series with $n=2$, so we already know it converges, but apply the ratio test nonetheless

All the terms here are positive, so we won't need the absolute value sign when taking the limit

Write and simplify the ratio $\frac{a_{n+1}}{a_n}$

$\begin{array}{rcl}\frac{\frac{1}{{\left(n+1\right)}^2}}{{\displaystyle \frac{1}{n^2}}}& =& \frac{1}{{\left(n+1\right)}^2}·\frac{n^2}{1}\\ & =& \frac{n^2}{{\left(n+1\right)}^2}\\ & =& {\left(\frac{n}{n+1}\right)}^2\\ & =& {\left(\frac{n}{n+1}·\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}}\right)}^2\\ & =& {\left(\frac{1}{1+{\displaystyle \frac{1}{n}}}\right)}^2\end{array}$

Take the limit

$\begin{array}{rcl}\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}& =& \underset{n\to \infty }{\lim }{\left(\frac{1}{1+{\displaystyle \frac{1}{n}}}\right)}^2\\ & =& {\left(\frac{1}{1+0}\right)}^2\\ & =& 1\end{array}$

The limit is equal to 1, so the test is inconclusive

Note that the ratio test would similarly fail for the divergent harmonic series $\sum _{n=1}^{\infty }\frac{1}{n}$

The ratio test cannot determine whether this series converges or diverges