009C Sample Midterm 2, Problem 1

Evaluate:

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

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

Foundations:
L'Hopital's Rule
Sum formula for geometric 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 $\ln y=-4,$ $y=e^{-4}.$
Now, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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 }{\big (}{\frac {n-4}{n}}{\big )}^{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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_1=\frac{1}{2}.}
Hence, the sum of the series is
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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}}

Final Answer:
(a)     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{4}}
(b)     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{2}{3}}

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