# 009C Sample Final 2, Problem 1

Test if the following sequences converge or diverge. Also find the limit of each convergent sequence.

(a)  $a_{n}={\frac {\ln(n)}{\ln(n+1)}}$ (b)  $a_{n}={\bigg (}{\frac {n}{n+1}}{\bigg )}^{n}$ Foundations:
L'Hopital'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:
First, we notice that  $\lim _{n\rightarrow \infty }{\frac {\ln(n)}{\ln(n+1)}}$ has the form  ${\frac {\infty }{\infty }}.$ So, we can use L'Hopital's Rule. To begin, we write
$\lim _{n\rightarrow \infty }{\frac {\ln(n)}{\ln(n+1)}}=\lim _{x\rightarrow \infty }{\frac {\ln(x)}{\ln(x+1)}}.$ Step 2:
Now, using L'Hopital's rule, we get
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\frac {\ln(n)}{\ln(n+1)}}}&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {\frac {1}{x}}{\frac {1}{x+1}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {x+1}{x}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }1+{\frac {1}{x}}}\\&&\\&=&\displaystyle {1.}\end{array}}$ (b)

Step 1:
Let

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

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

${\begin{array}{rcl}\displaystyle {\ln y}&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {\ln {\bigg (}{\frac {n}{n+1}}{\bigg )}}{\frac {1}{n}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {\ln {\bigg (}{\frac {x}{x+1}}{\bigg )}}{\frac {1}{x}}}}\\&&\\&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {{\frac {x+1}{x}}{\frac {1}{(x+1)^{2}}}}{{\big (}-{\frac {1}{x^{2}}}{\big )}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {-x^{2}}{x(x+1)}}}\\&&\\&=&\displaystyle {-1.}\end{array}}$ Step 4:
Since  $\ln y=-1,$ we know
$y=e^{-1}.$ (a)     $1$ (b)     $e^{-1}$ 