Finding the sum of series $\sum \limits_{n=1}^{\infty} (-1)^{n}\frac{n^2}{2^n}$ I have some problems in finding the values of series that follow this pattern: 
$$\sum \limits_{n=0}^{\infty} (-1)^{n}*..$$
For example: I have to find the value of this series 
$$\sum \limits_{n=1}^{\infty} (-1)^{n}\frac{n^2}{2^n}$$
Can you give me some tips on how I should calculate the value of this kind of series? Thank you.
 A: $$\sum_{n=1}^\infty(-1)^n\frac{n^2}{2^n}=\sum_{n=1}^\infty n^2\left(-\frac12\right)^n$$
Now,
$$\sum_{n=0}^\infty r^n=\frac1{1-r}$$ for $|r|<1$
Differentiate either sides to get $$\sum_{n=1}^\infty nr^{n-1}=\frac1{(1-r)^2}$$
Multiplying by $r,$
$$\implies\sum_{n=1}^\infty nr^n=\frac r{(1-r)^2}$$
Again Differentiate either sides and multiply by $r$
Set $r=-\dfrac12$
A: Hint: start with
$$f(x)=\sum_{n=0}^\infty(-x)^n=\cdots$$
and calculate
$$f'(x)=\sum_{n=0}^\infty n(-x)^{n-1}(-1)=\cdots$$
$$f''(x)=\cdots$$
A: Hints:
We recognize $\sum_{n=1}^{\infty}n^{2}z^{n}$ where $z=-\frac{1}{2}$.
Wellknown is $\left(1-z\right)^{-1}=\sum_{n=0}^{\infty}z^{n}$ under
condition $\left|z\right|<1$. 
What happens if you differentiate (twice)
on both sides? 
A: Let $f(x):=\sum_{n\ge 1}n^2x^n$ denotes the formal power series for $n^2$. Then you need to evaluate $f(-1/2).$ Now, let $g(x)=\sum_{n\ge 1}nx^n$. Then, $f(x)-g(x)=\sum_{n\ge 1}n(n-1)x^n=x^2\sum_{n\ge 2}n(n-1)x^{n-2}$Now, we know the formal power series$$\sum_{n\ge 0}x^n=\frac{1}{1-x}\\\Rightarrow \sum_{n\ge 1}nx^{n-1}=\frac{1}{(1-x)^2} \\ \Rightarrow \sum_{n\ge 2}n(n-1)x^{n-2}=\frac{2}{(1-x)^3}$$ So, $$g(x)=\frac{x}{(1-x)^2}\\ f(x)-g(x)=\frac{2x^2}{(1-x)^3}\\ \Rightarrow f(x)=\frac{x+x^2}{(1-x)^3}$$
