009C Sample Final 2, Problem 1 Detailed Solution

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

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

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

Final Answer:
(a)     $1$
(b)     $e^{-1}$

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