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=\frac1{\sqrt2}`.

It should be clear to you that
`\frac1{\sqrt2}<1`, 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}*\frac1{1-\frac1{\sqrt2}}=\frac1{\sqrt2-1}=1+\sqrt2`