# 009C Sample Midterm 1, Problem 1

Does the following sequence converge or diverge?

If the sequence converges, also find the limit of the sequence.

${\displaystyle a_{n}={\frac {\ln n}{n}}}$

Foundations:
L'Hôpital's Rule, Part 2

Let  ${\displaystyle f}$  and  ${\displaystyle g}$  be differentiable functions on the open interval  ${\displaystyle (a,\infty )}$  for some value  ${\displaystyle a,}$

where  ${\displaystyle g'(x)\neq 0}$  on  ${\displaystyle (a,\infty )}$  and  ${\displaystyle \lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}}$  returns either  ${\displaystyle {\frac {0}{0}}}$  or  ${\displaystyle {\frac {\infty }{\infty }}.}$
Then,   ${\displaystyle \lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}=\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}.}$

Solution:

Step 1:
First, notice that
${\displaystyle \lim _{n\rightarrow \infty }\ln n=\infty }$
and
${\displaystyle \lim _{n\rightarrow \infty }n=\infty .}$
Therefore, the limit has the form  ${\displaystyle {\frac {\infty }{\infty }},}$
which means that we can use L'Hopital's Rule to calculate this limit.
Step 2:
First, switch to the variable  ${\displaystyle x}$   so that we have functions and
can take derivatives. Thus, using L'Hopital's Rule, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\frac {\ln n}{n}}}&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {\ln x}{x}}}\\&&\\&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {{\big (}{\frac {1}{x}}{\big )}}{1}}}\\&&\\&=&\displaystyle {0.}\end{array}}}$

The sequence converges. The limit of the sequence is  ${\displaystyle 0.}$