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.
We have that
$$\sum_{n=1}^\infty 2^{-\frac{n}{2}}=\sum_{n=1}^\infty \left(\frac1{\sqrt 2}\right)^n$$
then refer to the geometric series.
$$\frac{1}{\sqrt2}+\frac{1}{2}+\frac{1}{2\sqrt2}+\frac{1}{4}+\frac{1}{4\sqrt2}+\frac18+\dots$$
$$=\dfrac1{\sqrt2}\left(1+\dfrac12+\dfrac14+...\right)+\left(\dfrac12+\dfrac14+\dfrac18+...\right)$$
$$=\dfrac1{\sqrt2}2+1=\sqrt2+1$$
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.
Note that, if $\lvert x\rvert<1$, then$$\sum_{n=1}^\infty x^n=\frac x{1-x}.$$Applying this with $x=2^{-1/2}=\frac1{\sqrt2}$, you get that the sum of your series is$$\frac{\frac1{\sqrt2}}{1-\frac1{\sqrt2}}=\frac1{\sqrt2-1}=\sqrt2+1.$$
