Series such as \[ S = 1 + \frac{2}{1!}+\frac{2^2}{2!}+\frac{2^3}{3!}+\frac{2^4}{4!}+\cdots \] arise in applications, but the convergence tests developed so far cannot be applied easily. Fortunately, the Ratio Test can be used for this and many other series.
Assume that the following limit exists: \[ \rho = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| \]
The symbol \(\rho\) is a lowercase “rho,” the seventeenth letter of the Greek alphabet.
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,
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\[ \begin{align*} |a_{M+1}|&< r|a_M|\\ |a_{M+2}|&< r|a_{M+1}| < r(r|a_M|) = r^2|a_M|\\ |a_{M+3}|&< r |a_{M+2} |< r^3|a_M| \end{align*} \]
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.
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.
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.
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Does \(\displaystyle\sum^\infty_{n=0} (-1)^{n}\frac{n!}{1000^n}\) converge?
Solution This series diverges by the Ratio Test because \(\rho>1\): \[ \rho = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\frac{(n+1)!}{1000^{n+1}}\,\frac{1000^n}{n!} = \lim_{n\to\infty}\frac{n+1}{1000}=\infty \]
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 \(\rho = 1\).
Our next test is based on the limit of the \(n\)th roots \(\sqrt[n]{a_n}\) rather than the ratios \(a_{n+1}/a_n\). Its proof, like that of the Ratio Test, is based on a comparison with a geometric series (see Exercise 57).
Assume that the following limit exists: \[ L=\lim_{n\to\infty} \sqrt[n]{|a_n|} \]
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.
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