009C Sample Final 1, Problem 3 Detailed Solution

Determine whether the following series converges or diverges.

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 \sum_{n=0}^{\infty} (-1)^n \frac{n!}{n^n}}

---

Background Information:
Recall:
1. Ratio Test Let 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 \sum a_n} be a series and 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 L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.} Then,
If 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 L<1,} the series is absolutely convergent.
If 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 L>1,} the series is divergent.
If 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 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
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}\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
$\lim _{n\rightarrow \infty }{\frac {\ln {\bigg (}{\frac {n}{n+1}}{\bigg )}}{\frac {1}{n}}}.$
Now, we use L'Hopital's Rule to get
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }n\ln {\bigg (}{\frac {n}{n+1}}{\bigg )}}&{\overset {l'H}{=}}&\displaystyle {\lim _{n\rightarrow \infty }{\frac {{\frac {n+1}{n}}{\frac {(n+1)-n}{(n+1)^{2}}}}{-{\frac {1}{n^{2}}}}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {1}{n(n+1)}}(-n^{2})}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {-n}{n+1}}}\\&&\\&=&\displaystyle {-1.}\\\end{array}}$

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.

Return to Sample Exam