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:
Review Ratio Test

Solution:

Step 1:
We proceed using the ratio test.
We have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{n\rightarrow \infty }n\ln {\bigg (}{\frac {n}{n+1}}{\bigg )}} .
First, we write the limit as Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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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