# 009C Sample Midterm 2, Problem 1 Detailed Solution

Evaluate:

(a)  $\lim _{n\rightarrow \infty }{\frac {1}{{\big (}{\frac {n-4}{n}}{\big )}^{n}}}$ (b)  $\sum _{n=1}^{\infty }{\frac {1}{2}}{\bigg (}{\frac {1}{4}}{\bigg )}^{n-1}$ Background Information:
1. L'Hôpital's Rule, Part 2

Let  $f$ and  $g$ be differentiable functions on the open interval  $(a,\infty )$ for some value  $a,$ where  $g'(x)\neq 0$ on  $(a,\infty )$ and  $\lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}$ returns either  ${\frac {0}{0}}$ or  ${\frac {\infty }{\infty }}.$ Then,   $\lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}=\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}.$ 2. The sum of a convergent geometric series is   ${\frac {a}{1-r}}$ where  $r$ is the ratio of the geometric series
and  $a$ is the first term of the series.

Solution:

(a)

Step 1:
Let

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

${\begin{array}{rcl}\displaystyle {\ln y}&=&\displaystyle {\lim _{n\rightarrow \infty }\ln {\bigg (}1-{\frac {4}{n}}{\bigg )}^{n}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }n\ln {\bigg (}1-{\frac {4}{n}}{\bigg )}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {\ln {\bigg (}1-{\frac {4}{n}}{\bigg )}}{\frac {1}{n}}}.}\end{array}}$ Now, this limit has the form  ${\frac {0}{0}}.$ Hence, we can use L'Hopital's Rule to calculate this limit.
Step 3:
Now, we have

${\begin{array}{rcl}\displaystyle {\ln y}&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {\ln {\bigg (}1-{\frac {4}{n}}{\bigg )}}{\frac {1}{n}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {\ln {\bigg (}1-{\frac {4}{x}}{\bigg )}}{\frac {1}{x}}}}\\&&\\&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {{\frac {1}{{\big (}1-{\frac {4}{x}}{\big )}}}{\frac {4}{x^{2}}}}{{\big (}-{\frac {1}{x^{2}}}{\big )}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {-4x}{x-4}}}\\&&\\&=&\displaystyle {-4.}\end{array}}$ Step 4:
Since  $\ln y=-4,$ we know
$y=e^{-4}.$ Now, we have

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

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

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