009C Sample Final 1, Problem 4

Find the interval of convergence of the following series.

$\sum _{n=0}^{\infty }(-1)^{n}{\frac {(x+2)^{n}}{n^{2}}}$

Foundations:
Recall:
1. Ratio Test Let $\sum a_{n}$ be a series and $L=\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}$. Then,
If $L<1$, the series is absolutely convergent.
If $L>1$, the series is divergent.
If Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L=1} , the test is inconclusive.
2. After you find the radius of convergence, you need to check the endpoints of your interval
for convergence since the Ratio Test is inconclusive when Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L=1} .

Solution:

Step 1:
We proceed using the ratio test to find the interval of convergence. So, we have
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(-1)^{n+1}(x+2)^{n+1}}{(n+1)^{2}}}}{\frac {n^{2}}{(-1)^{n}(x+2)^{n}}}{\bigg |}\\&&\\&=&\displaystyle {|x+2|\lim _{n\rightarrow \infty }{\frac {n^{2}}{(n+1)^{2}}}}\\&&\\&=&\displaystyle {|x+2|\lim _{n\rightarrow \infty }{\bigg (}{\frac {n}{n+1}}{\bigg )}^{2}}\\&&\\&=&\displaystyle {|x+2|{\bigg (}\lim _{n\rightarrow \infty }{\frac {n}{n+1}}{\bigg )}^{2}}\\&&\\&=&\displaystyle {|x+2|(1)^{2}}\\&&\\&=&\displaystyle {|x+2|.}\\\end{array}}$

Step 2:
So, we have $|x+2|<1$. Hence, our interval is $(-3,-1)$. But, we still need to check the endpoints of this interval
to see if they are included in the interval of convergence.

Step 3:
First, we let $x=-1$. Then, our series becomes $\sum _{n=0}^{\infty }(-1)^{n}{\frac {1}{n^{2}}}.$
Since $n^{2}<(n+1)^{2}$, we have ${\frac {1}{(n+1)^{2}}}<{\frac {1}{n^{2}}}.$ Thus, ${\frac {1}{n^{2}}}$ is decreasing.
So, $\sum _{n=0}^{\infty }(-1)^{n}{\frac {1}{n^{2}}}$ converges by the Alternating Series Test.

Step 4:
Now, we 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 x=-3} . Then, our series becomes
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{\sum_{n=0}^{\infty} (-1)^n \frac{(-1)^n}{n^2}} & = & \displaystyle{\sum_{n=0}^{\infty} (-1)^{2n} \frac{1}{n^2}}\\ &&\\ & = & \displaystyle{\sum_{n=0}^{\infty} \frac{1}{n^2}.}\\ \end{array}}
This is a convergent series by the p-test.

Step 5:
Thus, the interval of convergence for this series is 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 [-3,-1].}

Final Answer:
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 [-3,-1]}

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