009C Sample Final 2, Problem 1

Test if the following sequences converge or diverge. Also find the limit of each convergent sequence.

(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 a_n=\frac{\ln(n)}{\ln(n+1)}}

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

Foundations:
L'Hopital'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:
First, we notice that  $\lim _{n\rightarrow \infty }{\frac {\ln(n)}{\ln(n+1)}}$  has the form  ${\frac {\infty }{\infty }}.$
So, we can use L'Hopital's Rule. To begin, we write
$\lim _{n\rightarrow \infty }{\frac {\ln(n)}{\ln(n+1)}}=\lim _{x\rightarrow \infty }{\frac {\ln(x)}{\ln(x+1)}}.$

Step 2:
Now, using L'Hopital's rule, we get
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\frac {\ln(n)}{\ln(n+1)}}}&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {{\big (}{\frac {1}{x}}{\big )}}{{\big (}{\frac {1}{x+1}}{\big )}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {x+1}{x}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }1+{\frac {1}{x}}}\\&&\\&=&\displaystyle {1.}\end{array}}$

(b)

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

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

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

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 1}
(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 e^{-1}}

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