Question

{(1/sqrt(2))^n} from n=1 to infinity; is the series convergent or divergent? if convergent, why? and sum?

Answer 1

This is a geometric series with common ratio between terms of $r=\dfrac1{\sqrt2}$.

It should be clear to you that $\dfrac1{\sqrt2}$, which means the series will converge.

The sum would be

$\displaystyle\sum_{n\ge1}\left(\frac1{\sqrt2}\right)^n=\frac1{\sqrt2}\sum_{n\ge1}\left(\frac1{\sqrt2}\right)^{n-1}=\frac1{\sqrt2}\times\frac1{1-\frac1{\sqrt2}}=\frac1{\sqrt2-1}=1+\sqrt2$