009C Sample Midterm 1, Problem 1 Detailed Solution

Does the following sequence converge or diverge?

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

Be sure to jusify your answers!

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

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Background Information:
L'Hôpital's Rule, Part 2
Let  $f$  and  $g$  be differentiable functions on the open interval  $(a,\infty )$  for some value  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle a,}
where  $g'(x)\neq 0$  on  $(a,\infty )$  and  $\lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}$  returns either  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {0}{0}}}   or  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {\infty }{\infty }}.}
Then,   $\lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}=\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}.$

Solution:

Step 1:
First, notice that
$\lim _{n\rightarrow \infty }\ln n=\infty$
and
$\lim _{n\rightarrow \infty }n=\infty .$
Therefore, the limit has the form  ${\frac {\infty }{\infty }},$
which means that we can use L'Hopital's Rule to calculate this limit.

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

Final Answer:
The sequence converges. The limit of the sequence is  $0.$

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