find interval of convergence for series Is it right that the range of convergence is here $1 &lt; x &lt; 3$:

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

Just like you do with the geometric series? Or what is this radius of convergence? Thanks!

update:

i got until now:

$$\frac{\frac{(e^{n+1}+e^{-(n+1)})*(x-2)^{n+1}}{(n+1)^2}}{\frac{(e^n+e^{-n})*(x-2)^n}{n^2}}$$

(the middle fractal line should be the main one)
and this should be less the 1

right?
Using the Ratio Test, we have

$\displaystyle\lim_{n\to\infty}\frac{e^{n+1}+e^{-(n+1)}}{(n+1)^2}\lvert x-2\rvert^{n+1}\cdot\frac{n^2}{(e^n+e^{-n})\lvert x-2\rvert^n}=\lim_{n\to\infty}\frac{e^{n+1}+e^{-(n+1)}}{e^n+e^{-n}}\cdot\frac{n^2}{(n+1)^2}\cdot\lvert x-2\rvert$

$\displaystyle=\lim_{n\to\infty}\frac{e+e^{(-2n+1)}}{1+e^{-2n}}\cdot\left(\frac{n}{n+1}\right)^2\cdot\lvert x-2\rvert=e\cdot 1\cdot\lvert x-2\rvert=e\lvert x-2\rvert$,

and $e\lvert x-2\rvert&lt;1 \iff \lvert x-2\rvert&lt;\frac{1}{e}$.



To test convergence at the endpoints of the interval,

A) $\;\displaystyle x=2+\frac{1}{e}$ gives the series $\displaystyle\sum_{n=1}^{\infty}\frac{1+e^{-2n}}{n^2}$, 

$\;\;\;$which converges by comparing to $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}$ using the Limit Comparison Test.

B) $\;\displaystyle x=2-\frac{1}{e}$ gives the series
$\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{1+e^{-2n}}{n^2}$,

$\;\;\;$which converges since its absolute value converges.

Therefore the series converges for x in $\displaystyle\left[2-\frac{1}{e},2+\frac{1}{e}\right]$.
According to WolframAlpha, the interval of convergence is as follows.



You can see it plotted as so:



Unfortunately I'm not sure how one would go about finding it, but for the bottom  part you can use the p-series test and because it is n^2, p>2 so it is convergent everywhere, now you just have to decide for the top part what test to use.