Use of differentiation to find the power series and determine the sum of the series We were asked to differentiate $f(x) = \frac{1}{4-{x^2}}$ to determine the sum of the series $\sum_{n=1}^\infty \frac{n}{16^{n}}$
so far I got the power series representation which is
$$\sum_{n=0}^\infty \frac{x^{2n}}{4^{n+1}}$$
But when I differentiated the power series, my prof said it was wrong. The differentiated $f(x)$ and power series I got was:
$$f'(x) = \frac{2x}{(4-{x^2})^2}$$
$$\sum_{n=1}^\infty \frac{nx^{2n-1}}{2^{2n+1}}$$
How can I differentiate the power series and determine the sum of the series $\sum_{n=1}^\infty \frac{n}{16^{n}}$?
 A: From $$f(x)=\frac{1}{4-x^2}=\frac{1}{4}\frac{1}{1-\left(\frac{x}{2}\right)^2}=\frac{1}{4}\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^{2n}$$
Now
$$f'(x)=\frac{1}{4}\sum_{n=0}^{\infty}n\left(\frac{x}{2}\right)^{2n-1}=\frac{1}{4}\frac{2}{x}\sum_{n=0}^{\infty}n\left(\frac{x}{2}\right)^{2n}$$
Put $x=\frac{1}{2}$ and from the derivative you got you should get $\sum_{n=1}^{\infty}\frac{n}{16^n}=\frac{16}{225}$
A: Hint: I think it was a misunderstanding of what your professor meant. Your calculation is fine. On the one hand your calculated series evaluated at $x=\frac{1}{2}$ is
\begin{align*}
\sum_{n=1}^\infty \frac{nx^{2n-1}}{2^{2n+1}}\Big|_{x=\frac{1}{2}}
&=\sum_{n=1}^{\infty}\frac{n}{2^{2n+1}}\,\frac{1}{2^{2n-1}}=\sum_{n=1}^{\infty}\frac{n}{2^{4n}}=\color{blue}{\sum_{n=1}^{\infty}\frac{n}{16^n}}
\end{align*}
On the other hand you get
\begin{align*}
f^{\prime}\left(\frac{1}{2}\right)=\frac{1}{\left(4-\frac{1}{4}\right)^2}\color{blue}{=\frac{16}{225}}
\end{align*}
in accordance with the already given answer.
