Examples
Example 1:
\sum_{n=1}^{\infty} \frac{1}{2^n}
\begin{align*}
a_n &= \frac{1}{2^n} \\
a_{n+1} &= \frac{1}{2^{n+1}} \\
L &= \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{1}{2^{n+1}}}{\frac{1}{2^n}} \right| \\
&= \lim_{n \to \infty} \left| \frac{2^n}{2^{n+1}} \right| = \lim_{n \to \infty} \left| \frac{1}{2} \right| \\
&= \frac{1}{2}
\end{align*}
Since
L = \frac{1}{2} < 1
, the series
\sum_{n=1}^{\infty} \frac{1}{2^n}
converges by the Ratio Test.
Example 2: \sum_{n=1}^{\infty} 2^n
\begin{align*}
a_n &= 2^n \\
a_{n+1} &= 2^{n+1} \\
L &= \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{2^{n+1}}{2^n} \right| \\
&= \lim_{n \to \infty} \left| 2 \right| \\
&= 2
\end{align*}
Since
L = 2 > 1
, the series
\sum_{n=1}^{\infty} 2^n
diverges by the Ratio Test.
Example 3: \sum_{n=1}^{\infty} \frac{1}{n!}
\begin{align*}
a_n &= \frac{1}{n!} \\
a_{n+1} &= \frac{1}{(n+1)!} \\
L &= \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{1}{(n+1)!}}{\frac{1}{n!}} \right| \\
&= \lim_{n \to \infty} \left| \frac{n!}{(n+1)!} \right| = \lim_{n \to \infty} \left| \frac{1}{n+1} \right| \\
&= 0
\end{align*}
Since
L = 0 < 1
, the series
\sum_{n=1}^{\infty} \frac{1}{n!}
converges by the Ratio Test.
Example 4: Series with nn and Factorial - Divergent \sum_{n=1}^{\infty} \frac{n^n}{n!}
\begin{align*}
a_n &= \frac{n^n}{n!} \\
a_{n+1} &= \frac{(n+1)^{n+1}}{(n+1)!} \\
L &= \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{(n+1)^{n+1}}{(n+1)!}}{\frac{n^n}{n!}} \right| \\
&= \lim_{n \to \infty} \left| \frac{(n+1)^{n+1} \cdot n!}{(n+1)! \cdot n^n} \right| = \lim_{n \to \infty} \left| \frac{(n+1)^n \cdot (n+1) \cdot n!}{(n+1) \cdot n! \cdot n^n} \right| \\
&= \lim_{n \to \infty} \left| \frac{(n+1)^n}{n^n} \right| = \lim_{n \to \infty} \left| \left(\frac{n+1}{n}\right)^n \right| \\
&= \lim_{n \to \infty} \left| \left(1 + \frac{1}{n}\right)^n \right| = e
\end{align*}
Since
L = e > 1
, the series
\sum_{n=1}^{\infty} \frac{n^n}{n!}
diverges by the Ratio Test.
Example 5: \sum_{n=1}^{\infty} \frac{1}{n^2}
\begin{align*}
a_n &= \frac{1}{n^2} \\
a_{n+1} &= \frac{1}{(n+1)^2} \\
L &= \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{1}{(n+1)^2}}{\frac{1}{n^2}} \right| \\
&= \lim_{n \to \infty} \left| \frac{n^2}{(n+1)^2} \right| = \lim_{n \to \infty} \left| \left(\frac{n}{n+1}\right)^2 \right| \\
&= \lim_{n \to \infty} \left| \left(\frac{1}{1 + \frac{1}{n}}\right)^2 \right| = \left(\frac{1}{1 + 0}\right)^2 = 1
\end{align*}
Since
L = 1
, the Ratio Test is inconclusive for the series
\sum_{n=1}^{\infty} \frac{1}{n^2}
.
We know from the p-series test that this series converges.