<s>[INST] 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.
 [/INST] 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 $(3)$.
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