THEOREM 1 Ratio Test

Assume that the following limit exists: \[ \rho = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| \]

1. If \(\rho <1\), then \(\displaystyle\sum a_n\) converges absolutely.
2. If \(\rho >1\), then \(\displaystyle\sum a_n\) diverges.
3. If \(\rho =1\), the test is inconclusive (the series may converge or diverge).

The symbol \(\rho\) is a lowercase “rho,” the seventeenth letter of the Greek alphabet.

Proof

The idea is to compare with a geometric series. If \(\rho<1\), we may choose a number \(r\) such that \(\rho<r<1\). Since \(\left|{a_{n+1}}/{a_n}\right|\) converges to \(\rho\), there exists a number \(M\) such that \(\left|{a_{n+1}}/{a_n}\right|<r\) for all \(n\ge M\). Therefore,

\[ \begin{align*} |a_{M+1}|&< r|a_M|\\ |a_{M+2}|&< r|a_{M+1}| < r(r|a_M|) = r^2|a_M|\\ |a_{M+3}|&< r |a_{M+2} |< r^3|a_M| \end{align*} \]

In general, \(|a_{M+n}|<r^n|a_M|\), and thus, \[ \sum_{n=M}^\infty |a_n | = \sum_{n=0}^\infty |a_{M+n}| \le \sum_{n=0}^\infty |a_M|\, r^n = |a_M| \sum_{n=0}^\infty r^n \] The geometric series on the right converges because \(0< r <1\), so \({\displaystyle\sum_{n=M}^\infty |a_n| }\) converges by the Comparison Test and thus \({\displaystyle\sum a_n}\) converges absolutely.

If \(\rho>1\), choose \(r\) such that \(1<r<\rho\). Then there exists a number \(M\) such that \(\left| {a_{n+1}}/{a_n}\right|>r\) for all \(n\ge M\). Arguing as before with the inequalities reversed, we find that \(|a_{M+n}|\ge r^n|a_M|\). Since \(r^n\) tends to \(\infty\), the terms \(a_{M+n}\) do not tend to zero, and consequently, \({\displaystyle\sum a_n}\) diverges. Finally, Example 4 below shows that both convergence and divergence are possible when \(\rho=1\), so the test is inconclusive in this case.

EXAMPLE 1

Prove that \(\displaystyle\sum^\infty_{n=1} \frac{2^n}{n!}\) converges.

Solution Compute the ratio and its limit with \(a_n = \dfrac{2^n}{n!}\). Note that \((n+1)!=(n+1)n!\) and thus \[ \begin{eqnarray*} \frac{a_{n+1}}{a_n} &=&\frac{2^{n+1}}{(n+1)!}\frac{n!}{2^n} = \frac{2^{n+1}}{2^n}\frac{n!}{(n+1)!} = \frac2{n+1}\\ \rho &=& \lim_{n\to\infty} \left\vert\frac{a_{n+1}}{a_n}\right\vert = \lim_{n\to\infty} \frac2{n+1} = 0 \end{eqnarray*} \] Since \(\rho<1\), the series \(\displaystyle\sum^\infty_{n=1} \frac{2^n}{n!}\) converges by the Ratio Test.

EXAMPLE 2

Does \(\displaystyle\sum^\infty_{n=1} \frac{n^2}{2^n}\) converge?

Solution Apply the Ratio Test with \(a_n = \frac{n^2}{2^n}\): \[ \begin{eqnarray*} \left\vert\frac{a_{n+1}}{a_n}\right\vert &=&\frac{(n+1)^2}{2^{n+1}}\, \frac{2^n}{n^2} = \frac12\left(\frac{n^2+2n+1}{n^2}\right) = \frac12\left(1+\frac2n+\frac1{n^2}\right)\\ \rho &=& \lim_{n\to\infty} \left\vert\frac{a_{n+1}}{a_n}\right\vert = \frac12\lim_{n\to\infty} \left(1+\frac2n+\frac1{n^2}\right) = \frac12 \end{eqnarray*} \] Since \(\rho<1\), the series converges by the Ratio Test.

EXAMPLE 3

Does \(\displaystyle\sum^\infty_{n=0} (-1)^{n}\frac{n!}{1000^n}\) converge?

Solution This series diverges by the Ratio Test because \(\rho>1\): \[ \rho = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\frac{(n+1)!}{1000^{n+1}}\,\frac{1000^n}{n!} = \lim_{n\to\infty}\frac{n+1}{1000}=\infty \]

EXAMPLE 4 Ratio Test Inconclusive

Show that both convergence and divergence are possible when \(\rho=1\) by considering \({\displaystyle\sum_{n=1}^\infty n^2}\) and \({\displaystyle\sum_{n=1}^\infty n^{-2}}\).

Solution For \(a_n = n^2\), we have \[ \rho= \lim_{n\to\infty} \left\vert\frac{a_{n+1}}{a_n}\right\vert = \lim_{n\to\infty} \frac{(n+1)^2}{n^2} = \lim_{n\to\infty} \frac{n^2+2n+1}{n^2} = \lim_{n\to\infty} \left(1+\frac2n+\frac1{n^2}\right) = 1 \] On the other hand, for \(b_n=n^{-2}\), \[ \rho= \lim_{n\to\infty} \left\vert\frac{b_{n+1}}{b_n}\right\vert = \lim_{n\to\infty} \left\vert\frac{a_n}{a_{n+1}}\right\vert = \frac{1}{\lim\limits_{n\to\infty} \left\vert\frac{a_{n+1}}{a_n}\right\vert}=1 \] Thus, \(\rho = 1\) in both cases, but \({\displaystyle\sum_{n=1}^\infty n^2}\) diverges and \({\displaystyle\sum_{n=1}^\infty n^{-2}}\) converges. This shows that both convergence and divergence are possible when \(\rho = 1\).

THEOREM 2 Root Test

Assume that the following limit exists: \[ L=\lim_{n\to\infty} \sqrt[n]{|a_n|} \]

1. If \(L <1\), then \(\displaystyle\sum a_n\) converges absolutely.
2. If \(L >1\), then \(\displaystyle\sum a_n\) diverges.
3. If \(L =1\), the test is inconclusive (the series may converge or diverge).

EXAMPLE 5

Does \(\displaystyle\sum^\infty_{n=1} \left(\frac{n}{2n+3}\right)^n\) converge?

Solution We have \( L = \lim\limits_{n\to\infty} \sqrt[n]{|a_n|} = \lim\limits_{n\to\infty} \frac{n}{2n+3} = \frac12\). Since \(L<1\), the series converges by the Root Test.