009C Sample Final 1, Problem 4 Detailed Solution

Find the interval of convergence of the following series.

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

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Background Information:
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
$\sum _{n=0}^{\infty }(-1)^{n}{\frac {1}{n^{2}}}.$
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
${\frac {1}{(n+1)^{2}}}<{\frac {1}{n^{2}}}.$
Thus,  ${\frac {1}{n^{2}}}$  is decreasing.
Also,
$\lim _{n\rightarrow \infty }{\frac {1}{n^{2}}}=0.$
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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