009C Sample Final 3, Problem 1

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

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

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

ExpandFoundations:
L'Hôpital's Rule
Suppose that  $\lim _{x\rightarrow \infty }f(x)$  and  $\lim _{x\rightarrow \infty }g(x)$  are both zero or both  $\pm \infty .$
If  $\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}$  is finite or  $\pm \infty ,$
then  $\lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}\,=\,\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}.$

Solution:

(a)

ExpandStep 1:
Let         ${\begin{array}{rcl}\displaystyle {y}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg (}1+{\frac {1}{2n}}{\bigg )}^{n}.}\end{array}}$
We then take the natural log of both sides to get
$\ln y=\ln {\bigg (}\lim _{n\rightarrow \infty }{\bigg (}1+{\frac {1}{2n}}{\bigg )}^{n}{\bigg )}.$

ExpandStep 2:
We can interchange limits and continuous functions.
Therefore, we have
${\begin{array}{rcl}\displaystyle {\ln y}&=&\displaystyle {\lim _{n\rightarrow \infty }\ln {\bigg (}1+{\frac {1}{2n}}{\bigg )}^{n}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }n\ln {\bigg (}1+{\frac {1}{2n}}{\bigg )}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {\ln {\bigg (}1+{\frac {1}{2n}}{\bigg )}}{\frac {1}{n}}}.}\end{array}}$
Now, this limit has the form  ${\frac {0}{0}}.$
Hence, we can use L'Hopital's Rule to calculate this limit.

ExpandStep 3:
Now, we have
${\begin{array}{rcl}\displaystyle {\ln y}&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {\ln {\bigg (}1+{\frac {1}{2n}}{\bigg )}}{\frac {1}{n}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {\ln {\bigg (}1+{\frac {1}{2x}}{\bigg )}}{\frac {1}{x}}}}\\&&\\&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {{\frac {2x}{2x+1}}{\big (}{\frac {-1}{2x^{2}}}{\big )}}{{\big (}-{\frac {1}{x^{2}}}{\big )}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {1}{2}}{\bigg (}{\frac {2x}{2x+1}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {1}{2}}.}\end{array}}$

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

(b)

ExpandStep 1:
First, we have
$\lim _{n\rightarrow \infty }\cos(n\pi ){\bigg (}{\frac {1+n}{n}}{\bigg )}^{n}=\lim _{n\rightarrow \infty }(-1)^{n}{\bigg (}{\frac {1+n}{n}}{\bigg )}^{n}.$

ExpandStep 2:
Now, let
${\begin{array}{rcl}\displaystyle {y}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg (}{\frac {1+n}{n}}{\bigg )}^{n}.}\end{array}}$
We then take the natural log of both sides to get
$\ln y=\ln {\bigg (}\lim _{n\rightarrow \infty }{\bigg (}{\frac {1+n}{n}}{\bigg )}^{n}{\bigg )}.$
We can interchange limits and continuous functions.
Therefore, we have
${\begin{array}{rcl}\displaystyle {\ln y}&=&\displaystyle {\lim _{n\rightarrow \infty }\ln {\bigg (}{\frac {1+n}{n}}{\bigg )}^{n}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }n\ln {\bigg (}{\frac {1+n}{n}}{\bigg )}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {\ln {\bigg (}{\frac {1+n}{n}}{\bigg )}}{\frac {1}{n}}}.}\end{array}}$
Now, this limit has the form  ${\frac {0}{0}}.$
Hence, we can use L'Hopital's Rule to calculate this limit.

ExpandStep 3:
Now, we have
${\begin{array}{rcl}\displaystyle {\ln y}&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {\ln {\bigg (}{\frac {1+n}{n}}{\bigg )}}{\frac {1}{n}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {\ln {\bigg (}{\frac {1+x}{x}}{\bigg )}}{\frac {1}{x}}}}\\&&\\&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {{\frac {x}{1+x}}{\big (}{\frac {-1}{x^{2}}}{\big )}}{{\big (}-{\frac {1}{x^{2}}}{\big )}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {x}{1+x}}}\\&&\\&=&\displaystyle {1.}\end{array}}$

ExpandStep 4:
Since  $\ln y=1,$  we know
$y=e.$
Since
$\lim _{n\rightarrow \infty }{\bigg (}{\frac {1+n}{n}}{\bigg )}^{n}\neq 0,$
we have
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }a_{n}}&=&\displaystyle {\lim _{n\rightarrow \infty }(-1)^{n}{\bigg (}{\frac {1+n}{n}}{\bigg )}^{n}}\\&&\\&=&\displaystyle {{\text{DNE}}.}\end{array}}$

ExpandFinal Answer:
(a)    $e^{1/2}$
(b)    ${\text{DNE}}$

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