009C Sample Final 2, Problem 1

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

(a)  $a_{n}={\frac {\ln(n)}{\ln(n+1)}}$

(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
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}\frac{\ln(n)}{\ln(n+1)}} & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{x}}{\frac{1}{x+1}}}\\ &&\\ & = & \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         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{n}{n+1}\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{n}{n+1}\bigg)^n\bigg).}

Step 2:
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{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  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{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  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^{-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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