009C Sample Final 1, Problem 3

Determine whether the following series converges or diverges.

$\sum _{n=0}^{\infty }(-1)^{n}{\frac {n!}{n^{n}}}$

Foundations:
Recall:
1. Ratio Test Let $\sum a_{n}$ be a series and $L=\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}$. Then,
If $L<1$, the series is absolutely convergent.
If $L>1$, the series is divergent.
If $L=1$, the test is inconclusive.
2. If a series absolutely converges, then it also converges.

Solution:

Step 1:
We proceed using the ratio test.
We have
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(-1)^{n+1}(n+1)!}{(n+1)^{n+1}}}{\frac {n^{n}}{(-1)^{n}n!}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(n+1)n!}{n!}}{\frac {n^{n}}{(n+1)^{n+1}}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(n+1)n^{n}}{(n+1)(n+1)^{n}}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\bigg (}{\frac {n}{n+1}}{\bigg )}^{n}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg (}{\frac {n}{n+1}}{\bigg )}^{n}.}\\\end{array}}$

Step 2:
Now, we continue to calculate the limit from Step 1. We have
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg (}{\frac {n}{n+1}}{\bigg )}^{n}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }e^{\ln({\frac {n}{n+1}})^{n}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }e^{n\ln({\frac {n}{n+1}})}}\\&&\\&=&\displaystyle {e^{\lim _{n\rightarrow \infty }n\ln({\frac {n}{n+1}})}.}\\\end{array}}$

Step 3:
Now, we need to calculate $\lim _{n\rightarrow \infty }n\ln {\bigg (}{\frac {n}{n+1}}{\bigg )}.$
First, we write the limit as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \rightarrow \infty}\frac{\ln\bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}.}
Now, we use L'Hopital's Rule to get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\lim_{n \rightarrow \infty}n\ln\bigg(\frac{n}{n+1}\bigg)} & \overset{l'H}{=} & \displaystyle{\lim_{n \rightarrow \infty}\frac{\frac{n+1}{n}\frac{(n+1)-n}{(n+1)^2}}{-\frac{1}{n^2}}}\\ &&\\ & = & \displaystyle{\lim_{n \rightarrow \infty} \frac{1}{n(n+1)}(-n^2)}\\ &&\\ & = & \displaystyle{\lim_{n \rightarrow \infty} \frac{-n}{n+1}}\\ &&\\ & = & \displaystyle{-1.}\\ \end{array}}

Step 4:
We go back to Step 2 and use the limit we calculated in Step 3.
So, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|=e^{-1}=\frac{1}{e}<1.}
Thus, the series absolutely converges by the Ratio Test.
Since the series absolutely converges, the series also converges.

Final Answer:
The series converges.

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