009C Sample Final 2, Problem 1 Detailed Solution

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

(a)  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle a_{n}={\frac {\ln(n)}{\ln(n+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 a_n=\bigg(\frac{n}{n+1}\bigg)^n}

Background Information:  
L'Hopital's Rule

        Suppose that  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_{x\rightarrow \infty} f(x)}   and  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_{x\rightarrow \infty} g(x)}   are both zero or both   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 \pm \infty .}

       If  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_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}}   is finite or   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 \pm \infty ,}

       then  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_{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  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} \frac{\ln(n)}{\ln(n+1)}}   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{\infty}{\infty}.}
So, we can use L'Hopital's Rule. To begin, we write
       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}\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:  
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}\cdot \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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