Find $\sum\limits_{n=1}^{\infty}\frac{n^2}{2^n}$ using the function $f(x)=\frac{1+x}{(1-x)^3}$ Find $\sum\limits_{n=1}^{\infty}\frac{n^2}{2^n}$ using the function $f(x)=\frac{1+x}{(1-x)^3}$
Power series representation of $f(x)$ is $\sum\limits_{n=1}^{\infty}n^2x^{n-1}$.
Question: Why is $\sum\limits_{n=1}^{\infty}\frac{n^2}{2^n}=\frac{1}{2}f\left(\frac{1}{2}\right)$? Doesn't it has to be $f\left(\frac{1}{2}\right)$?

You need the extra factor of $2$:
$$f(1/2) = \sum_{n=1}^{\infty}\frac{n^2}{2^{n-1}},$$
but
$$\frac{1}{2}f(1/2) = \sum_{n=1}^{\infty}\frac{n^2}{2 \cdot 2^{n-1}} = \sum_{n=1}^{\infty}\frac{n^2}{2^{n}}.$$
