Evaluate $\sum_{n=1}^\infty 2^{-\frac{n}{2}}$ 
Find $$\sum_{n=1}^\infty 2^{-\frac{n}{2}}$$

I know that the final numerical value of that is $1+\sqrt2$ but not sure how to get that. Any identities, formula or hints would be helpful. 
I tried re-expressing it as $\frac{1}{\sqrt2}+\frac{1}{2}+\frac{1}{2\sqrt2}+\frac{1}{4}+\frac{1}{4\sqrt2}+\dots$ but it doesn't seem useful.
The closest looking formula I can find is $\sum_{n=0}^\infty x^n=\frac{1}{1-x}$ but even that doesn't seem to apply for this.

You have $$\sum_{n=1}^\infty 2^{-\frac n2}=\sum_{n=1}^\infty (2^{-\frac 12})^n = \frac{2^{-\frac{1}2}}{1-2^{-\frac{1}2}} = 1+\sqrt 2$$
by the same formula you gave.
