Determine the sum of function series The serie is : 

$$\sum _{n=1}^{\infty }{\frac { \left( -1 \right) ^{n} \left( 2\,n-1
 \right) {x}^{2\,n}}{{4}^{n}}}
$$

I found out the derivative $2\,{\frac {n \left( -1 \right) ^{n}{\left( 2\,n-1 \right)}{x}^{2\,n-1}
}{{4}^{n}}}
$ , which gave me this $2\,n-1$ in power and in numerator, but I have no idea what to do next. So what do I do next ? Any detailed explanation would be very appreciated.

$$\sum _{n=1}^{\infty }{\frac { \left( -1 \right) ^{n} \left( 2\,n-1
 \right) {x}^{2\,n}}{{4}^{n}}}=\sum _{n=1}^{\infty } \frac{(-1)^n (2 n) x^{2 n}}{4^n} + \sum _{n=1}^{\infty } \frac{(-1)^{n+1} x^{2 n}}{4^n}$$
For the first series
$$\sum _{n=1}^{\infty } \frac{(-1)^n  x^{2 n}}{4^n}=\sum _{n=1}^{\infty } \left(-\frac{x^2}{4}\right)^n  =\frac{1}{1+x^2/4}-1=-\frac{x^2}{x^2+4}$$
Then derive and get
$$\sum _{n=1}^{\infty } \frac{(-1)^n (2 n) x^{2 n}}{4^n} =-\frac{8 x^2}{\left(x^2+4\right)^2}$$
The second series is
$$\sum _{n=1}^{\infty } \frac{(-1)^{n+1} x^{2 n}}{4^n}=-\sum _{n=1}^{\infty } \left(-\frac{x^2}{4}\right)^{n} =-\left(\frac{1}{1+\frac{x^2}{4}}-1\right)=\frac{x^2}{x^2+4}$$
so the sum is
$$-\frac{8 x^2}{\left(x^2+4\right)^2}+\frac{x^2}{x^2+4}=\frac{x^2 \left(x^2-4\right)}{\left(x^2+4\right)^2}$$
Hope this is useful
