# Supremum of series

#### Fermat

##### Active member
$Sup(\sum_{k=n+1}^{\infty}\frac{|x_{k}|^{2}}{4^{k}})$ where
$x=(x_{1},x_{2},....)$ is in $l_{2}$ and the supremum is taken over all $x$ such that $||x||$=1.

I think it is equal to $\frac{1}{4^{n+1}}$ Is this correct?