# 009C Sample Final 1, Problem 4

Find the interval of convergence of the following series.

${\displaystyle \sum _{n=0}^{\infty }(-1)^{n}{\frac {(x+2)^{n}}{n^{2}}}}$
Foundations:
Recall:
1. Ratio Test Let ${\displaystyle \sum a_{n}}$ be a series and ${\displaystyle L=\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}.}$ Then,
If ${\displaystyle L<1,}$ the series is absolutely convergent.
If ${\displaystyle L>1,}$ the series is divergent.
If ${\displaystyle L=1,}$ 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 ${\displaystyle L=1.}$

Solution:

Step 1:
We proceed using the ratio test to find the interval of convergence. So, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(-1)^{n+1}(x+2)^{n+1}}{(n+1)^{2}}}}{\frac {n^{2}}{(-1)^{n}(x+2)^{n}}}{\bigg |}\\&&\\&=&\displaystyle {|x+2|\lim _{n\rightarrow \infty }{\frac {n^{2}}{(n+1)^{2}}}}\\&&\\&=&\displaystyle {|x+2|\lim _{n\rightarrow \infty }{\bigg (}{\frac {n}{n+1}}{\bigg )}^{2}}\\&&\\&=&\displaystyle {|x+2|{\bigg (}\lim _{n\rightarrow \infty }{\frac {n}{n+1}}{\bigg )}^{2}}\\&&\\&=&\displaystyle {|x+2|(1)^{2}}\\&&\\&=&\displaystyle {|x+2|.}\\\end{array}}}$
Step 2:
So, we have ${\displaystyle |x+2|<1.}$ Hence, our interval is ${\displaystyle (-3,-1).}$ 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 ${\displaystyle x=-1.}$ Then, our series becomes
${\displaystyle \sum _{n=0}^{\infty }(-1)^{n}{\frac {1}{n^{2}}}.}$
Since ${\displaystyle n^{2}<(n+1)^{2},}$ we have ${\displaystyle {\frac {1}{(n+1)^{2}}}<{\frac {1}{n^{2}}}.}$ Thus, ${\displaystyle {\frac {1}{n^{2}}}}$ is decreasing.
So, ${\displaystyle \sum _{n=0}^{\infty }(-1)^{n}{\frac {1}{n^{2}}}}$ converges by the Alternating Series Test.
Step 4:
Now, we let ${\displaystyle x=-3.}$ Then, our series becomes
${\displaystyle {\begin{array}{rcl}\displaystyle {\sum _{n=0}^{\infty }(-1)^{n}{\frac {(-1)^{n}}{n^{2}}}}&=&\displaystyle {\sum _{n=0}^{\infty }(-1)^{2n}{\frac {1}{n^{2}}}}\\&&\\&=&\displaystyle {\sum _{n=0}^{\infty }{\frac {1}{n^{2}}}.}\\\end{array}}}$
This is a convergent series by the p-test.
Step 5:
Thus, the interval of convergence for this series is ${\displaystyle [-3,-1].}$
${\displaystyle [-3,-1]}$