In this section we learn two tests for convergence of series: the Root Test and the Ratio Test.
Ratio Test: Consider a series \(\displaystyle\sum_{n=1}^{\infty} a_n\). Let
\[L=\lim_{n\to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert.\]
If \(L<1\), then \(\displaystyle\sum_{n=1}^{\infty} a_n\) is absolutely convergent.
If \(L>1\), then \(\displaystyle\sum_{n=1}^{\infty} a_n\) is divergent.
If \(L=1\), then this test is inconclusive.
(a) Suppose \(L < 1\). Let \(r\) be a number such that \(L < r < 1\). Since
\(\displaystyle\lim_{n\to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert=L < r\),
there is a positive integer \(K\) such that
\[ \left\vert \frac{a_{n+1}}{a_n} \right\vert < r, \text{ i.e., } |a_{n+1}| < r|a_n| \text{ for all } n\geq K. \]
Successively applying the last inequality, we get
\[ |a_{K+n}| < r^n|a_K| \text{ for all } n\geq 1. \]
Since \(\displaystyle \sum_{n=1}^{\infty} r^n|a_K|\) is convergent as a geometric series,
\(\displaystyle \sum_{n=1}^{\infty} |a_{K+n}|\) is also convergent by the Comparison Test. Then so is the
following as the sum of a finite number and a convergent series:
\[\sum_{n=1}^{\infty} |a_n| = \sum_{n=1}^{K} |a_n| +\sum_{n=1}^{\infty} |a_{K+n}|.\]
Thus \(\displaystyle\sum_{n=1}^{\infty} a_n\) is absolutely convergent.
(b) Suppose \(\displaystyle\lim_{n\to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert=L>1\). Then there is
a positive integer \(K\) such that
\[ \left\vert \frac{a_{n+1}}{a_n} \right\vert>1, \text{ i.e., } |a_{n+1}| > |a_n| \text{ for all } n\geq K. \]
Then \(\displaystyle\lim_{n\to \infty} a_n\neq 0\). Thus \(\displaystyle\sum_{n=1}^{\infty} a_n\) is divergent
by the Divergence Test.
(c) When \(L=1\), \(\displaystyle\sum_{n=1}^{\infty} a_n\) may be divergent (e.g., \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}\)),
absolutely convergent (e.g., \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}\)), and
conditionally convergent (e.g., \(\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\)).
Example.
Use the Ratio Test to determine if \(\displaystyle \sum_{n=1}^{\infty} \frac{n!}{n^n}\) is
absolutely convergent or divergent.
Solution. Here \(a_n=\displaystyle\frac{n!}{n^n}\). Then \(a_{n+1}=\displaystyle\frac{(n+1)!}{(n+1)^{n+1}}\) and
\[\begin{align*}
L &=\lim_{n\to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert \\
&=\lim_{n\to \infty} \left\vert \frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}} \right\vert \\
&=\lim_{n\to \infty} \frac{(n+1)!}{(n+1)^{n+1}} \frac{n^n}{n!} \\
&=\lim_{n\to \infty} \frac{(n+1)\cdot n!}{(n+1)\cdot (n+1)^n} \frac{n^n}{n!} \\
&=\lim_{n\to \infty} \left( \frac{n}{n+1} \right)^n\\
&= \frac{1}{\displaystyle\lim_{n\to \infty} \left(1+\frac{1}{n}\right)^n} \\
&= \frac{1}{e}.
\end{align*}\]
Since \(L=\frac{1}{e}<1\), \(\displaystyle \sum_{n=1}^{\infty} \frac{n!}{n^n}\) is absolutely convergent by
the Ratio Test and hence convergent.
Use the Ratio Test to determine if \(\displaystyle \sum_{n=1}^{\infty} \frac{2^n}{n^2}\) is
absolutely convergent or divergent.
Solution. Here \(a_n=\displaystyle\frac{2^n}{n^2}\). Then \(a_{n+1}=\displaystyle\frac{2^{n+1}}{(n+1)^2}\) and
\[\begin{align*}
L &=\lim_{n\to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert \\
&=\lim_{n\to \infty} \left\vert \frac{\frac{2^{n+1}}{(n+1)^2}}{\frac{2^n}{n^2}} \right\vert \\
&=\lim_{n\to \infty} \frac{2^{n+1}}{(n+1)^2} \frac{n^2}{2^n} \\
&=\lim_{n\to \infty} \frac{2\cdot 2^n}{2^n} \left(\frac{n}{n+1}\right)^2 \\
&=\lim_{n\to \infty} 2 \left(\frac{1}{1+\frac{1}{n}}\right)^2\\
&= 2 \left(\lim_{n\to \infty}\frac{1}{1+\frac{1}{n}}\right)^2 \\
&= 2.
\end{align*}\]
Since \(L=2>1\), \(\displaystyle \sum_{n=1}^{\infty} \frac{2^n}{n^2}\) is divergent by the Ratio Test.
Use the Ratio Test to determine if \(\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}\) is
absolutely convergent or divergent.
Solution. Here \(a_n=\displaystyle \frac{1}{n^2}\). Then \(a_{n+1}=\displaystyle \frac{1}{(n+1)^2}\)
and
\[\begin{align*}
L &=\lim_{n\to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert \\
&=\lim_{n\to \infty} \frac{1}{(n+1)^2} \frac{n^2}{1} \\
&=\lim_{n\to \infty} \left(\frac{1}{1+\frac{1}{n}}\right)^2\\
&= \left(\lim_{n\to \infty}\frac{1}{1+\frac{1}{n}}\right)^2 \\
&= 1.
\end{align*}\]
Since \(L= 1\), the Ratio Test is inconclusive (although \(\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}\)
is convergent by the \(p\)-series Test). Similarly we can show for \(\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}\)
that \(L=1\) and the Ratio Test is inconclusive where the series is known to be divergent.
Root Test: Consider a series \(\displaystyle\sum_{n=1}^{\infty} a_n\). Let
\[L=\lim_{n\to \infty}\displaystyle \left\vert a_n \right\vert^{\frac{1}{n}}.\]
If \(L < 1\), then \(\displaystyle\sum_{n=1}^{\infty} a_n\) is absolutely convergent.
If \(L > 1\), then \(\displaystyle\sum_{n=1}^{\infty} a_n\) is divergent.
If \(L=1\), then this test is inconclusive.
The proof of the Root Test is similar to that of the Ratio Test.
Example.
Use the Root Test to determine if \(\displaystyle \sum_{n=1}^{\infty} \left( \frac{-2n}{n+1}\right)^{3n}\)
is absolutely convergent or divergent.
Solution.
Here \(a_n=\displaystyle (-1)^{3n}\left(\frac{2n}{n+1} \right)^{3n}\). Then \(|a_n|=\displaystyle \left(\frac{2n}{n+1} \right)^{3n}\)
and
\[\begin{align*}
L&=\lim_{n\to \infty} \left\vert a_n \right\vert^{\frac{1}{n}}\\
&=\lim_{n\to \infty} \left[ \left(\frac{2n}{n+1} \right)^{3n} \right]^{\frac{1}{n}} \\
&=\lim_{n\to \infty} \left(\frac{2n}{n+1} \right)^{3}\\
&= \left( \lim_{n\to \infty} \frac{2}{1+\frac{1}{n}} \right)^{3}\\
&= 2^3\\
&= 8.
\end{align*}\]
Since \(L=8 > 1\), \(\displaystyle \sum_{n=1}^{\infty} \left( \frac{-2n}{n+1}\right)^{3n}\) is divergent by
the Root Test.
Note that this problem can also be done by the Ratio Test but the calculation will be a little bit more
complicated:
Here \(a_n=\displaystyle (-2)^{3n}\left(\frac{n}{n+1} \right)^{3n}\). Then
\(a_{n+1}=\displaystyle(-2)^{3n+3}\left(\frac{n+1}{n+2} \right)^{3n+3}\) and
\[\begin{align*}
L &=\lim_{n\to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert \\
&=\lim_{n\to \infty} \left\vert (-2)^{3n+3}\left(\frac{n+1}{n+2} \right)^{3n+3} \frac{1}{(-2)^{3n}}\left(\frac{n+1}{n} \right)^{3n} \right\vert \\
&=\lim_{n\to \infty} \left\vert -8\left(\frac{1+\frac{1}{n}}{1+\frac{2}{n}} \right)^{3n+3} \left(1+\frac{1}{n} \right)^{3n} \right\vert \\
&=\lim_{n\to \infty} 8\frac{\left(1+\frac{1}{n}\right)^{3}\left[\left(1+\frac{1}{n} \right)^{n}\right]^3}{\left(1+\frac{2}{n}\right)^{3}\left[\left(1+\frac{2}{n} \right)^{n}\right]^3} \left[\left(1+\frac{1}{n} \right)^{n}\right]^3 \\
&= 8 \frac{1^3\cdot e^3}{1^3\cdot (e^2)^3} e^3 \;\;\left( \text{since } \lim_{n\to \infty} \left(1+\frac{x}{n}\right)^n=e^x \right) \\
&= 8.
\end{align*}\]
Since \(L=8 > 1\), \(\displaystyle \sum_{n=1}^{\infty} \left( \frac{-2n}{n+1}\right)^{3n}\) is divergent by
the Ratio Test.