# 009C Sample Midterm 2, Problem 3 Detailed Solution

Determine convergence or divergence:

(a)  ${\displaystyle \sum _{n=1}^{\infty }(-1)^{n}{\sqrt {\frac {1}{n}}}}$

(b)  ${\displaystyle \sum _{n=1}^{\infty }(-2)^{n}{\frac {n!}{n^{n}}}}$

Background Information:
1. Alternating Series Test
Let  ${\displaystyle \{a_{n}\}}$  be a positive, decreasing sequence where  ${\displaystyle \lim _{n\rightarrow \infty }a_{n}=0.}$
Then,  ${\displaystyle \sum _{n=1}^{\infty }(-1)^{n}a_{n}}$  and  ${\displaystyle \sum _{n=1}^{\infty }(-1)^{n+1}a_{n}}$
converge.
2. 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.

3. If a series absolutely converges, then it also converges.

Solution:

(a)

Step 1:
First, we have
${\displaystyle \sum _{n=1}^{\infty }(-1)^{n}{\sqrt {\frac {1}{n}}}=\sum _{n=1}^{\infty }(-1)^{n}{\frac {1}{\sqrt {n}}}.}$
Step 2:
We notice that the series is alternating.
Let  ${\displaystyle b_{n}={\frac {1}{\sqrt {n}}}.}$
First, we have
${\displaystyle {\frac {1}{\sqrt {n}}}\geq 0}$
for all  ${\displaystyle n\geq 1.}$
The sequence  ${\displaystyle \{b_{n}\}}$  is decreasing since
${\displaystyle {\frac {1}{\sqrt {n+1}}}<{\frac {1}{\sqrt {n}}}}$
for all  ${\displaystyle n\geq 1.}$
Also,
${\displaystyle \lim _{n\rightarrow \infty }b_{n}=\lim _{n\rightarrow \infty }{\frac {1}{\sqrt {n}}}=0.}$
Therefore, the series  ${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{\sqrt {n}}}}$  converges by the Alternating Series Test.

(b)

Step 1:
We begin by using the Ratio Test.
We have

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(-2)^{n+1}(n+1)!}{(n+1)^{n+1}}}{\frac {n^{n}}{(-2)^{n}n!}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}(-2)(n+1){\frac {n^{n}}{(n+1)^{n+1}}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }2{\frac {n^{n}}{(n+1)^{n}}}}\\&&\\&=&\displaystyle {2\lim _{n\rightarrow \infty }{\bigg (}{\frac {n}{n+1}}{\bigg )}^{n}.}\end{array}}}$

Step 2:
Now, we need to calculate  ${\displaystyle \lim _{n\rightarrow \infty }{\bigg (}{\frac {n}{n+1}}{\bigg )}^{n}.}$
Let  ${\displaystyle y=\lim _{n\rightarrow \infty }{\bigg (}{\frac {n}{n+1}}{\bigg )}^{n}.}$
Then, taking the natural log of both sides, we get

${\displaystyle {\begin{array}{rcl}\displaystyle {\ln y}&=&\displaystyle {\ln {\bigg (}\lim _{n\rightarrow \infty }{\bigg (}{\frac {n}{n+1}}{\bigg )}^{n}{\bigg )}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }\ln {\bigg (}{\frac {n}{n+1}}{\bigg )}^{n}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }n\ln {\bigg (}{\frac {n}{n+1}}{\bigg )}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {\ln {\bigg (}{\frac {n}{n+1}}{\bigg )}}{\frac {1}{n}}}}\end{array}}}$

since we can interchange limits and continuous functions.
Now, this limit has the form  ${\displaystyle {\frac {0}{0}}.}$
Hence, we can use L'Hopital's Rule to calculate this limit.
Step 3:
Now, we have

${\displaystyle {\begin{array}{rcl}\displaystyle {\ln y}&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {\ln {\big (}{\frac {n}{n+1}}{\big )}}{\frac {1}{n}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {\ln {\big (}{\frac {x}{x+1}}{\big )}}{\frac {1}{x}}}}\\&&\\&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {{\frac {1}{{\big (}{\frac {x}{x+1}}{\big )}}}{\frac {1}{(x+1)^{2}}}}{{\big (}-{\frac {1}{x^{2}}}{\big )}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {-x}{x+1}}}\\&&\\&=&\displaystyle {-1.}\end{array}}}$

Step 4:
Since  ${\displaystyle \ln y=-1,}$  we know
${\displaystyle y=e^{-1}.}$
Now, we have
${\displaystyle \lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}=2e^{-1}={\frac {2}{e}}.}$
Since  ${\displaystyle {\frac {2}{e}}<1,}$  the series is absolutely convergent by the Ratio Test.
Therefore, the series converges.