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

(b)

Step 1:
First, we not that this is a geometric series with  $r={\frac {1}{4}}.$
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
${\begin{array}{rcl}\displaystyle {\frac {a_{1}}{1-r}}&=&\displaystyle {\frac {\frac {1}{2}}{1-{\frac {1}{4}}}}\\&&\\&=&\displaystyle {\frac {{\big (}{\frac {1}{2}}{\big )}}{{\big (}{\frac {3}{4}}{\big )}}}\\&&\\&=&\displaystyle {{\frac {2}{3}}.}\end{array}}$

Final Answer:
(a)     $e^{4}$
(b)     ${\frac {2}{3}}$

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