Calculate the sum of $\sum_{n=1}^\infty (-1)^{n-1} \frac{n+1}{2^\frac{n}{2}}$ So, we have $$\sum_{n=1}^\infty (-1)^{n-1} \frac{n+1}{2^\frac{n}{2}}$$
I immediately noticed that $ln(1+x) = \sum_{n=1}^\infty (-1)^{n-1} \frac{x^n}{n}$ , for $|x|\lt 1$
It reminds of this sum, if we switch nominator and denominator and multiply it by $2^{n/2+1}$, we would get something similar.
Thats all i got so far.
 A: One may start with the standard finite evaluation:
$$
1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1. \tag1
$$ Then by differentiating $(1)$ we get
$$
1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x}, \quad |x|<1, \tag2
$$ by making $n \to +\infty$ in $(2)$, using $|x|<1$, gives 
$$
\sum_{n=0}^\infty(n+1)x^n=\frac{1}{(1-x)^2}. \tag3
$$
Now just insert $x:=-\dfrac1{\sqrt{2}}$ into $(3)$.
A: You are on the right track thinking about power series.
Note that whenever you see something like $$f(x) = \sum_{n=1}^\infty (n+1) a_n x^n$$ then an anti-derivative is given by $$F(x) = C + \sum_{n=1}^\infty a_n x^{n+1}$$ for some constant $C$.
Thus $$\sum_{n=1}^\infty (-1)^{n-1} (n+1) x^n$$ has an anti-derivative of the form $$F(x)=\sum_{n=1}^\infty (-1)^{n-1} x^{n+1} = (-1) x \sum_{n=1}^\infty (-1)^n x^n = (-x)\cdot \frac{-x}{1+x}.$$
Finally, the sum you seek is given by $F'(1/\sqrt{2})$
A: HINT:
$$\begin{align}
(-1)^{n-1}\frac{n+1}{2^{n/2}}&=- (n+1)\left(-\frac{1}{\sqrt{2}}\right)^{n}\\\\
&=-\left.\frac{d}{dx}\left(x^{n+1}\right)\right|_{x=-\frac{1}{\sqrt{2}}}
\end{align}$$
And the series $\sum_{n=1}^\infty (n+1)x^n$ converges uniformly for $|x|\le r<1$.
