# 009C Sample Midterm 2, Problem 1 Detailed Solution

Evaluate:

(a)  ${\displaystyle \lim _{n\rightarrow \infty }{\frac {1}{{\big (}{\frac {n-4}{n}}{\big )}^{n}}}}$

(b)  ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2}}{\bigg (}{\frac {1}{4}}{\bigg )}^{n-1}}$

Background Information:
1. 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,   ${\displaystyle \lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}=\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}.}$
2. The sum of a convergent geometric series is   ${\displaystyle {\frac {a}{1-r}}}$
where  ${\displaystyle r}$  is the ratio of the geometric series
and  ${\displaystyle a}$  is the first term of the series.

Solution:

(a)

Step 1:
Let

${\displaystyle {\begin{array}{rcl}\displaystyle {y}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg (}{\frac {n-4}{n}}{\bigg )}^{n}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg (}1-{\frac {4}{n}}{\bigg )}^{n}.}\end{array}}}$

We then take the natural log of both sides to get
${\displaystyle \ln y=\ln {\bigg (}\lim _{n\rightarrow \infty }{\bigg (}1-{\frac {4}{n}}{\bigg )}^{n}{\bigg )}.}$
Step 2:
We can interchange limits and continuous functions.
Therefore, we have

${\displaystyle {\begin{array}{rcl}\displaystyle {\ln y}&=&\displaystyle {\lim _{n\rightarrow \infty }\ln {\bigg (}1-{\frac {4}{n}}{\bigg )}^{n}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }n\ln {\bigg (}1-{\frac {4}{n}}{\bigg )}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {\ln {\bigg (}1-{\frac {4}{n}}{\bigg )}}{\frac {1}{n}}}.}\end{array}}}$

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

${\displaystyle {\begin{array}{rcl}\displaystyle {\ln y}&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {\ln {\bigg (}1-{\frac {4}{n}}{\bigg )}}{\frac {1}{n}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {\ln {\bigg (}1-{\frac {4}{x}}{\bigg )}}{\frac {1}{x}}}}\\&&\\&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {{\frac {1}{{\big (}1-{\frac {4}{x}}{\big )}}}{\frac {4}{x^{2}}}}{{\big (}-{\frac {1}{x^{2}}}{\big )}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {-4x}{x-4}}}\\&&\\&=&\displaystyle {-4.}\end{array}}}$

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

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\frac {1}{{\big (}{\frac {n-4}{n}}{\big )}^{n}}}}&=&\displaystyle {\frac {\displaystyle {\lim _{n\rightarrow \infty }1}}{\displaystyle {\lim _{n\rightarrow \infty }{\bigg (}{\frac {n-4}{n}}{\bigg )}^{n}}}}\\&&\\&=&\displaystyle {\frac {1}{e^{-4}}}\\&&\\&=&\displaystyle {e^{4}.}\end{array}}}$

(b)

Step 1:
First, we not that this is a geometric series with  ${\displaystyle r={\frac {1}{4}}.}$
Since  ${\displaystyle |r|={\frac {1}{4}}<1,}$
this series converges.
Step 2:
Now, we need to find the sum of this series.
The first term of the series is  ${\displaystyle a_{1}={\frac {1}{2}}.}$
Hence, the sum of the series is

${\displaystyle {\begin{array}{rcl}\displaystyle {\frac {a_{1}}{1-r}}&=&\displaystyle {\frac {\frac {1}{2}}{1-{\frac {1}{4}}}}\\&&\\&=&\displaystyle {\frac {{\big (}{\frac {1}{2}}{\big )}}{{\big (}{\frac {3}{4}}{\big )}}}\\&&\\&=&\displaystyle {{\frac {2}{3}}.}\end{array}}}$

(a)     ${\displaystyle e^{4}}$
(b)     ${\displaystyle {\frac {2}{3}}}$