009C Sample Midterm 2, Problem 3 Detailed Solution

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Determine convergence or divergence:

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

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

Solution:

(a)

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

(b)

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

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

${\begin{array}{rcl}\displaystyle {\ln y}&=&\displaystyle {\ln {\bigg (}\lim _{n\rightarrow \infty }{\bigg (}{\frac {n}{n+1}}{\bigg )}^{n}{\bigg )}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }\ln {\bigg (}{\frac {n}{n+1}}{\bigg )}^{n}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }n\ln {\bigg (}{\frac {n}{n+1}}{\bigg )}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {\ln {\bigg (}{\frac {n}{n+1}}{\bigg )}}{\frac {1}{n}}}}\end{array}}$ since we can interchange limits and continuous functions.
Now, this limit has the form  ${\frac {0}{0}}.$ Hence, we can use L'Hopital's Rule to calculate this limit.
Step 3:
Now, we have

${\begin{array}{rcl}\displaystyle {\ln y}&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {\ln {\big (}{\frac {n}{n+1}}{\big )}}{\frac {1}{n}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {\ln {\big (}{\frac {x}{x+1}}{\big )}}{\frac {1}{x}}}}\\&&\\&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {{\frac {1}{{\big (}{\frac {x}{x+1}}{\big )}}}{\frac {1}{(x+1)^{2}}}}{{\big (}-{\frac {1}{x^{2}}}{\big )}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {-x}{x+1}}}\\&&\\&=&\displaystyle {-1.}\end{array}}$ Step 4:
Since  $\ln y=-1,$ we know
$y=e^{-1}.$ Now, we have
$\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}=2e^{-1}={\frac {2}{e}}.$ Since  ${\frac {2}{e}}<1,$ the series is absolutely convergent by the Ratio Test.
Therefore, the series converges.