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:
Ratio Test
Check endpoints of interval

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 $x=-3$. Then, our series becomes
${\begin{array}{rcl}\displaystyle {\sum _{n=0}^{\infty }(-1)^{n}{\frac {(-1)^{n}}{n^{2}}}}&=&\displaystyle {\sum _{n=0}^{\infty }(-1)^{2n}{\frac {1}{n^{2}}}}\\&&\\&=&\displaystyle {\sum _{n=0}^{\infty }{\frac {1}{n^{2}}}}\\\end{array}}$
This is a convergent series by the p-test.

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

Final Answer:
$[-3,-1]$

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