# 009C Sample Final 3, Problem 1

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Which of the following sequences  ${\displaystyle (a_{n})_{n\geq 1}}$  converges? Which diverges? Give reasons for your answers!

(a)  ${\displaystyle a_{n}={\bigg (}1+{\frac {1}{2n}}{\bigg )}^{n}}$

(b)  ${\displaystyle a_{n}=\cos(n\pi ){\bigg (}{\frac {1+n}{n}}{\bigg )}^{n}}$

Foundations:
L'Hôpital's Rule

Suppose that  ${\displaystyle \lim _{x\rightarrow \infty }f(x)}$  and  ${\displaystyle \lim _{x\rightarrow \infty }g(x)}$  are both zero or both  ${\displaystyle \pm \infty .}$

If  ${\displaystyle \lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}}$  is finite or  ${\displaystyle \pm \infty ,}$

then  ${\displaystyle \lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}\,=\,\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}.}$

Solution:

(a)

Step 1:
Let

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

We then take the natural log of both sides to get
${\displaystyle \ln y=\ln {\bigg (}\lim _{n\rightarrow \infty }{\bigg (}1+{\frac {1}{2n}}{\bigg )}^{n}{\bigg )}.}$
Step 2:
We can interchange limits and continuous functions.
Therefore, we have

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

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

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

Step 4:
Since  ${\displaystyle \ln y=1/2,}$  we know
${\displaystyle y=e^{1/2}.}$

(b)

Step 1:
First, we have
${\displaystyle \lim _{n\rightarrow \infty }\cos(n\pi ){\bigg (}{\frac {1+n}{n}}{\bigg )}^{n}=\lim _{n\rightarrow \infty }(-1)^{n}{\bigg (}{\frac {1+n}{n}}{\bigg )}^{n}.}$
Step 2:
Now, let
${\displaystyle {\begin{array}{rcl}\displaystyle {y}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg (}{\frac {1+n}{n}}{\bigg )}^{n}.}\end{array}}}$
We then take the natural log of both sides to get
${\displaystyle \ln y=\ln {\bigg (}\lim _{n\rightarrow \infty }{\bigg (}{\frac {1+n}{n}}{\bigg )}^{n}{\bigg )}.}$
We can interchange limits and continuous functions.
Therefore, we have

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

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

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

Step 4:
Since  ${\displaystyle \ln y=1,}$  we know
${\displaystyle y=e.}$
Since
${\displaystyle \lim _{n\rightarrow \infty }{\bigg (}{\frac {1+n}{n}}{\bigg )}^{n}\neq 0,}$
we have

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

(a)    ${\displaystyle e^{1/2}}$
(b)    ${\displaystyle {\text{DNE}}}$