Calculate the sum of $\sum_{n=1}^\infty (-1)^{n-1} \frac{n+1}{2^\frac{n}{2}}$

Question

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.

Answer 1

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 $(3)$.

Answer 2

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 $F'(1/\sqrt{2})$

Answer 3

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 $|x|\le r<1$.