2

Is it right that the range of convergence is here $1 < x < 3$:

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

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?

user3435407
  • 1,099

2 Answers2

1

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 $e\lvert x-2\rvert<1 \iff \lvert x-2\rvert<\frac{1}{e}$.

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 $\displaystyle\left[2-\frac{1}{e},2+\frac{1}{e}\right]$.

user84413
  • 27,211
  • Is $|x-2|\lt \frac 1e$ strictly necessary, or is it possible that we have $\le$? – abiessu Jun 02 '15 at 00:34
  • ok thanks! the absolute value I still don't understand, but i will think it through tomorrow morning, but now I can't even see any more... – user3435407 Jun 02 '15 at 00:35
  • @abiessu: is it like if we derive this analog from the geometric series, than it can't be equal to 1 cause then it would be a division by 0, right? – user3435407 Jun 02 '15 at 00:37
  • @abiessu Good question - I have only found the open interval of convergence, so I will edit my answer to test the endpoints. – user84413 Jun 02 '15 at 00:37
  • 1
    @user3435407: that would be true if we had only a geometric-type series, but we have the $1\over n^2$ portion as well... – abiessu Jun 02 '15 at 00:38
  • ok, got everything, thanks for the help – user3435407 Jun 02 '15 at 08:34
-1

According to WolframAlpha, the interval of convergence is as follows.

Radius of Convergence

You can see it plotted as so:

Plot of interval of convergence

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.

CyanogenCX
  • 155