# 009C Sample Final 1, Problem 3

Determine whether the following series converges or diverges.

${\displaystyle \sum _{n=0}^{\infty }(-1)^{n}{\frac {n!}{n^{n}}}}$
Foundations:
Recall:
1. Ratio Test Let ${\displaystyle \sum a_{n}}$ be a series and ${\displaystyle L=\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}.}$ Then,
If ${\displaystyle L<1,}$ the series is absolutely convergent.
If ${\displaystyle L>1,}$ the series is divergent.
If ${\displaystyle L=1,}$ 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
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(-1)^{n+1}(n+1)!}{(n+1)^{n+1}}}{\frac {n^{n}}{(-1)^{n}n!}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(n+1)n!}{n!}}{\frac {n^{n}}{(n+1)^{n+1}}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(n+1)n^{n}}{(n+1)(n+1)^{n}}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\bigg (}{\frac {n}{n+1}}{\bigg )}^{n}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg (}{\frac {n}{n+1}}{\bigg )}^{n}.}\\\end{array}}}$
Step 2:
Now, we continue to calculate the limit from Step 1. We have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg (}{\frac {n}{n+1}}{\bigg )}^{n}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }e^{\ln({\frac {n}{n+1}})^{n}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }e^{n\ln({\frac {n}{n+1}})}}\\&&\\&=&\displaystyle {e^{\lim _{n\rightarrow \infty }n\ln({\frac {n}{n+1}})}.}\\\end{array}}}$
Step 3:
Now, we need to calculate ${\displaystyle \lim _{n\rightarrow \infty }n\ln {\bigg (}{\frac {n}{n+1}}{\bigg )}.}$
First, we write the limit as
${\displaystyle \lim _{n\rightarrow \infty }{\frac {\ln {\bigg (}{\frac {n}{n+1}}{\bigg )}}{\frac {1}{n}}}.}$
Now, we use L'Hopital's Rule to get
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }n\ln {\bigg (}{\frac {n}{n+1}}{\bigg )}}&{\overset {l'H}{=}}&\displaystyle {\lim _{n\rightarrow \infty }{\frac {{\frac {n+1}{n}}{\frac {(n+1)-n}{(n+1)^{2}}}}{-{\frac {1}{n^{2}}}}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {1}{n(n+1)}}(-n^{2})}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {-n}{n+1}}}\\&&\\&=&\displaystyle {-1.}\\\end{array}}}$
Step 4:
We go back to Step 2 and use the limit we calculated in Step 3.
So, we have
${\displaystyle \lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}=e^{-1}={\frac {1}{e}}<1.}$
Thus, the series absolutely converges by the Ratio Test.
Since the series absolutely converges, the series also converges.