009C Sample Final 1, Problem 4

Find the interval of convergence of the following series.

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

Foundations:
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  $L=1,$  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   $L=1.$

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
$\sum _{n=1}^{\infty }(-1)^{n}{\frac {1}{n^{2}}}.$
We notice that this series is alternating.
Let  $b_{n}={\frac {1}{n^{2}}}.$
First, we have
${\frac {1}{n^{2}}}\geq 0$
for all  $n\geq 1.$
The sequence  $\{b_{n}\}$  is decreasing since
${\frac {1}{(n+1)^{2}}}<{\frac {1}{n^{2}}}$
for all  $n\geq 1.$
Also,
$\lim _{n\rightarrow \infty }b_{n}=\lim _{n\rightarrow \infty }{\frac {1}{n^{2}}}=0.$
So,  $\sum _{n=1}^{\infty }(-1)^{n}{\frac {1}{n^{2}}}$   converges by the Alternating Series Test.

Step 4:
Now, we let   $x=-3.$  Then, our series becomes
${\begin{array}{rcl}\displaystyle {\sum _{n=1}^{\infty }(-1)^{n}{\frac {(-1)^{n}}{n^{2}}}}&=&\displaystyle {\sum _{n=1}^{\infty }(-1)^{2n}{\frac {1}{n^{2}}}}\\&&\\&=&\displaystyle {\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}.}\\\end{array}}$
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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