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

Foundations:
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)

Step 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
$\ln y=\ln {\bigg (}\lim _{n\rightarrow \infty }{\bigg (}1+{\frac {1}{2n}}{\bigg )}^{n}{\bigg )}.$

Step 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.

(b)

Step 1:

Step 2:

Final Answer:
(a)
(b)

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