009C Sample Midterm 3, Problem 3 Detailed Solution

Test if each the following series converges or diverges.

Give reasons and clearly state if you are using any standard test.

(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 {\displaystyle \sum_{n=1}^{\infty}}\,\frac{n!}{(3n+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 {\displaystyle \sum_{n=2}^{\infty}}\,\frac{\sqrt{n}}{n^{2}-3}}

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Background Information:
1. Ratio Test
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 \sum a_n}   be a series 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 L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.}
Then,
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 L<1,}   the series is absolutely convergent.
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 L>1,}   the series is divergent.
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 L=1,}   the test is inconclusive.
2. If a series absolutely converges, then it also converges.
3. Limit Comparison Test
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 \{a_n\}}   and  $\{b_{n}\}$  be positive sequences.
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_{n\rightarrow \infty} \frac{a_n}{b_n}=L,}   where  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 L}  is a positive real number,
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 \sum_{n=1}^\infty a_n}   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 \sum_{n=1}^\infty b_n}   either both converge or both diverge.

Solution:

(a)

Step 1:
We begin by using the Ratio Test.
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}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(n+1)!}{(3(n+1)+1)!} \frac{(3n+1)!}{n!}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{(n+1)n!}{(3n+4)!} \frac{(3n+1)!}{n!}}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{(n+1)(3n+1)!}{(3n+4)(3n+3)(3n+2)(3n+1)!}}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n+1}{(3n+4)(3n+3)(3n+2)}}\\ &&\\ & = & \displaystyle{0.} \end{array}}

Step 2:
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 0<1,}   the series is absolutely convergent by the Ratio Test.
Therefore, the series converges.

(b)

Step 1:
First, we note 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 \frac{\sqrt{n}}{n^2-3}>0}
for all  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 n\ge 2.}
This means that we can use a comparison test on this series.
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 a_n=\frac{\sqrt{n}}{n^2-3}.}

Step 2:
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 b_n=\frac{\sqrt{n}}{n^2}=\frac{1}{n^{\frac{3}{2}}}.}
We want to compare the series in this problem with
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 \sum_{n=1}^\infty b_n=\sum_{n=2}^\infty \frac{1}{n^{\frac{3}{2}}}.}
This is 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 p} -series with  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 p=\frac{3}{2}.}
Hence,  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 \sum_{n=2}^\infty b_n}   converges.

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{\lim_{n\rightarrow \infty} \frac{a_n}{b_n}} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\big(\frac{\sqrt{n}}{n^2-3}\big)}{\bigg(\frac{1}{n^{\frac{3}{2}}}\bigg)}}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{(\sqrt{n})n^{\frac{3}{2}}}{n^2-3}}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n^2}{n^2-3}}\\ &&\\ & = & \displaystyle{1.} \end{array}}
Therefore, the series
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 \sum_{n=2}^{\infty} \frac{\sqrt{n}}{n^2-3}}
converges by the Limit Comparison Test.

Final Answer:
(a)     converges (by the Ratio Test)
(b)     converges (by the Limit Comparison Test)

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