This is a geometric series with common ratio between terms of [tex]r=\dfrac1{\sqrt2}[/tex].
It should be clear to you that [tex]\dfrac1{\sqrt2}<1[/tex], which means the series will converge.
The sum would be
[tex]\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[/tex]
