# 009C Sample Final 2, Problem 1

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

(a)  ${\displaystyle a_{n}={\frac {\ln(n)}{\ln(n+1)}}}$

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

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

(b)

Step 1:
Let

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

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

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

Now, this limit has the form  ${\displaystyle {\frac {0}{0}}.}$
Hence, we can use L'Hopital's Rule to calculate this limit.
Step 3:
Now, we have

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

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

(a)     ${\displaystyle 1}$
(b)     ${\displaystyle e^{-1}}$