# 009C Sample Midterm 3, Problem 1 Detailed Solution

Test if the following sequence ${\displaystyle {a_{n}}}$ converges or diverges.

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

${\displaystyle a_{n}=\left({\frac {n-7}{n}}\right)^{\frac {1}{n}}}$

Background Information:
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, let  ${\displaystyle L=\lim _{n\rightarrow \infty }\left({\frac {n-7}{n}}\right)^{1/n}.}$
Then,
${\displaystyle {\begin{array}{rcl}\ln L&=&\displaystyle {\ln \left(\lim _{n\rightarrow \infty }\left[\left({\frac {n-7}{n}}\right)^{1/n}\right]\right)}\\\\&=&\displaystyle {\lim _{n\rightarrow \infty }\ln \left[\left({\frac {n-7}{n}}\right)^{1/n}\right]}\\\\&=&\displaystyle {\lim _{n\rightarrow \infty }\left[{\frac {1}{n}}\cdot \ln \left({\frac {n-7}{n}}\right)\right]}\\\\&=&0\cdot \ln(1)\\\\&=&0.\end{array}}}$
Thus,  ${\displaystyle L=e^{0}=1.}$
Step 2:
Therefore, the sequence converges.
Additionally, the limit of the sequence is ${\displaystyle e^{0}=1.}$

The sequence converges. The limit of the sequence is  ${\displaystyle 1.}$