# Power Series Exercise

I'm having 2 problems at finding the correct interval of convergence of this power series.

$$\sum_{n=1}^∞ \frac{n(x-3)^{2n+1}}{2^{n}-1}$$

What I did was use the good old Ratio Test:

$\lim\limits_{n \to ∞} \left\lvert\frac{(n+1)(x-3)^{2n+3}}{2^{n+1}-1}*\frac{2^{n}-1}{n(x-3)^{2n+1}}\right\rvert$

Simplyfying the expression I got:

$\lim\limits_{n \to ∞} \left\lvert\frac{n+1}{n}*\frac{2^{n}-1}{2^{n+1}-1}*(x-3)^{2}\right\rvert$

The first limit equals $1$ and the second one equals $1/2$ so I get:

$\lim\limits_{n \to ∞} \left\lvert 1*\frac{1}{2}*(x-3)^{2}\right\rvert$=$\left\lvert\ \frac{(x-3)^{2}}{2}\right\rvert$

Since I want the Interval of $convergence$, my limit must be less than $1$:

$\left\lvert\ \frac{(x-3)^{2}}{2}\right\rvert$<$1$

Now here is the first problem, If I solve this, I get:

$-1$<$\frac{(x-3)^{2}}{2}$<$1$

$-2$<$(x-3)^{2}$<$2$

$\sqrt{-2}$<$x-3$<$\sqrt{2}$

$\sqrt{-2}+3$<$x$<$\sqrt{2}+3$

The lower value of the interval is not a real number at all, how do I evaluate this?.

The second problem comes when I evaluate the upper value of the interval $x=\sqrt{2}+3$. This is the Series that I obtain:

$$\sum_{n=1}^∞ \frac{n(\sqrt{2}+3-3)^{2n+1}}{2^{n}-1}$$

$$\sum_{n=1}^∞ \frac{n(\sqrt{2})^{2n+1}}{2^{n}-1}$$

$$\sum_{n=1}^∞ \frac{n(2^{1/2})^{2n+1}}{2^{n}-1}$$

$$\sum_{n=1}^∞ \frac{n(2)^{n+1/2}}{2^{n}-1}$$

I look at this Series and I just can't tell if it converges or diverges, It doesn't look like anything I did before.

So, basically that's it, I can't tell what happens with the lower value of the interval of convergence and I can't find a way to prove the Power series at $x=\sqrt{2}+3$ is Convergent or Divergent.

Hope it wasn't too long, I wanted to show everything I did.

Thank you for any help.

Following your steps, the convergence is ensured for all real numbers $x$ such that $$(x-3)^2<2 \iff |x-3|<\sqrt{2} \iff 3-\sqrt{2}<x<3+\sqrt{2}.$$
• By putting $x=3+\sqrt{2}$ into the series one has $$\sum_{n=1}^∞ \frac{n(x-3)^{2n+1}}{2^{n}-1}=\sqrt{2}\cdot\sum_{n=1}^∞ \frac{n\cdot2^{n}}{2^{n}-1}$$ the latter series does not converge since $$\lim_{n \to \infty}\frac{n\cdot2^{n}}{2^{n}-1}=\infty\ne0.$$
• By putting $x=3-\sqrt{2}$ into the series one has $$\sum_{n=1}^∞ \frac{n(x-3)^{2n+1}}{2^{n}-1}=-\sqrt{2}\cdot\sum_{n=1}^∞ \frac{n\cdot2^{n}}{2^{n}-1}$$ which does not converge as seen previously.