009C Sample Final 3, Problem 1

Which of the following sequences  $(a_{n})_{n\geq 1}$  converges? Which diverges? Give reasons for your answers!

(a)  $a_{n}={\bigg (}1+{\frac {1}{2n}}{\bigg )}^{n}$

(b)  $a_{n}=\cos(n\pi ){\bigg (}{\frac {1+n}{n}}{\bigg )}^{n}$

Foundations:
L'Hôpital's Rule
Suppose that  $\lim _{x\rightarrow \infty }f(x)$  and  $\lim _{x\rightarrow \infty }g(x)$  are both zero or both  $\pm \infty .$
If  $\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}$  is finite or  $\pm \infty ,$
then  $\lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}\,=\,\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}.$

Solution:

(a)

Step 1:
Let         ${\begin{array}{rcl}\displaystyle {y}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg (}1+{\frac {1}{2n}}{\bigg )}^{n}.}\end{array}}$
We then take the natural log of both sides to get
$\ln y=\ln {\bigg (}\lim _{n\rightarrow \infty }{\bigg (}1+{\frac {1}{2n}}{\bigg )}^{n}{\bigg )}.$

Step 2:
We can interchange limits and continuous functions.
Therefore, we have
${\begin{array}{rcl}\displaystyle {\ln y}&=&\displaystyle {\lim _{n\rightarrow \infty }\ln {\bigg (}1+{\frac {1}{2n}}{\bigg )}^{n}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }n\ln {\bigg (}1+{\frac {1}{2n}}{\bigg )}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {\ln {\bigg (}1+{\frac {1}{2n}}{\bigg )}}{\frac {1}{n}}}.}\end{array}}$
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
${\begin{array}{rcl}\displaystyle {\ln y}&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {\ln {\bigg (}1+{\frac {1}{2n}}{\bigg )}}{\frac {1}{n}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {\ln {\bigg (}1+{\frac {1}{2x}}{\bigg )}}{\frac {1}{x}}}}\\&&\\&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {{\frac {2x}{2x+1}}{\big (}{\frac {-1}{2x^{2}}}{\big )}}{{\big (}-{\frac {1}{x^{2}}}{\big )}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {1}{2}}{\bigg (}{\frac {2x}{2x+1}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {1}{2}}.}\end{array}}$

Step 4:
Since  $\ln y={\frac {1}{2}},$  we know
$y=e^{\frac {1}{2}}.$

(b)

Step 1:
First, we have
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 \lim_{n\rightarrow \infty}\cos(n\pi)\bigg(\frac{1+n}{n}\bigg)^n=\lim_{n\rightarrow \infty} (-1)^n\bigg(\frac{1+n}{n}\bigg)^n.}

Step 2:
Now, let
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{y} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{1+n}{n}\bigg)^n.} \end{array}}
We then take the natural log of both sides to get
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 \ln y = \ln\bigg(\lim_{n\rightarrow \infty} \bigg(\frac{1+n}{n}\bigg)^n\bigg).}
We can interchange limits and continuous functions.
Therefore, we have
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{\ln y} & = & \displaystyle{\lim_{n\rightarrow \infty} \ln \bigg(\frac{1+n}{n}\bigg)^n}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} n\ln\bigg(\frac{1+n}{n}\bigg)}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{1+n}{n}\bigg)}{\frac{1}{n}}.} \end{array}}
Now, this limit has the form  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{0}{0}.}
Hence, we can use L'Hopital's Rule to calculate this limit.

Step 3:
Now, we have
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{\ln y } & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{1+n}{n}\bigg)}{\frac{1}{n}}}\\ &&\\ & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(\frac{1+x}{x}\bigg)}{\frac{1}{x}}}\\ &&\\ & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{x}{1+x}\frac{-1}{x^2}}{\big(-\frac{1}{x^2}\big)}}\\ &&\\ & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{x}{1+x}}\\ &&\\ & = & \displaystyle{1.} \end{array}}

Step 4:
Since  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 \ln y= 1,}   we know
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 y=e.}
Since
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 \lim_{n\rightarrow \infty} \bigg(\frac{1+n}{n}\bigg)^n\neq 0,}
we have
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{\lim_{n\rightarrow \infty} a_n} & = & \displaystyle{\lim_{n\rightarrow \infty} (-1)^n\bigg(\frac{1+n}{n}\bigg)^n}\\ &&\\ & = & \displaystyle{\text{DNE}.} \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^{\frac{1}{2}}}
(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 \text{DNE}}

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