009C Sample Final 1, Problem 4

Find the interval of convergence of the following 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=0}^{\infty} (-1)^n \frac{(x+2)^n}{n^2}}

Foundations:
Ratio Test
Check endpoints of interval

Solution:

Step 1:
We proceed using the ratio test to find the interval of convergence. So, 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{(-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 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+2|<1} . Hence, our interval 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)} . 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 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=-1} . 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 sum_{n=0}^{\infty} (-1)^n \frac{1}{n^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 n^2<(n+1)^2} , 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 \frac{1}{(n+1)^2}<\frac{1}{n^2}} . Thus, 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{1}{n^2}} is decreasing.
So, 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=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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