Sum of power series $\sum_{n=1}^\infty (-1)^n\frac{n(n+1)}{2^n}x^n$ Calculate the sum of series:
$$\sum_{n=1}^\infty (-1)^n\frac{n(n+1)}{2^n}x^n$$ 
I know how to calculate sum of power series, but I don't know what should I do with $(-1)^n$
 A: Hint:
Try first to find what is 
$$
\sum_{n=1}^\infty n (n+1)y^n
$$
by noting that
$$
n(n+1)y^n=(y^{n+1})''\cdot y
$$
A most important technique about calculation of power series is differentiation (and integration) term by term, which should be discussed in any serious real (complex) analysis textbook. See also a note by Gowers. 

Besides the "formal" calculation, another issue you still need to address is that for what $y$ the series is convergent. 
A: HINT:
$$\sum_{n=1}^\infty (-1)^n\frac{n(n+1)}{2^n}x^n=\sum_{n=1}^\infty n^2\left(-\dfrac12\right)^n+\sum_{n=1}^\infty n\left(-\dfrac12\right)^n$$ 
Now for $|y|<1,$ $$\sum_{r=0}^\infty y^r=\dfrac1{1-y}$$
Differentiate both sides wrt $y$  $$\sum_{r=1}^\infty ry^{r-1}=\dfrac1{(1-y)^2}$$
$$\sum_{r=1}^\infty ry^r=\dfrac y{(1-y)^2}=\dfrac{1-(1-y)}{(1-y)^2}=\dfrac1{(1-y)^2}-\dfrac1{1-y}$$
Differentiate both sides wrt $y$ and multiply by $y$
A: $$\frac{1}{1+x}=\sum _{n=0}^{\infty }(-1)^{n}x^n$$
$$\frac{-x}{1+x}=\sum _{n=1}^{\infty }(-1)^{n}x^n$$
multiply by $x$
$$\frac{-x^2}{1+x}=\sum _{n=1}^{\infty }(-1)^{n}x^{n+1}$$
$$(\frac{-x^2}{1+x})''=\sum _{n=1}^{\infty }(-1)^{n}n(n+1)x^{n-1}$$
then multiply by $x$ and let $x\rightarrow \frac{x}{2}$
A: Let $y=-\dfrac{x}{2}$, so we have to find the power series 
$$\sum_{r=1}^{\infty}{r(r+1)y^r}$$
Note that
$$\sum_{r=1}^{\infty}{y^r}=\dfrac{1}{1-y}$$
On differentiating throughout twice we arrive at the following
$$\sum_{r=1}^{\infty}{r(r-1)y^{r-2}}=\dfrac{2y}{(1-y)^3}$$
From here onward it is pretty straightforward if you assign $r=n+1$ and substitute $y=-\dfrac{x}{2}$.
It is also to be noted that for the power series to converge $|y|\lt 1\implies \left|{x}\right|\lt 2$
