Supremum of $l_2$ Series: Is $\frac{1}{4^{n+1}}$ Correct?

[Fermat1]

$Sup(\sum_{k=n+1}^{\infty}\frac{|x_{k}|^{2}}{4^{k}})$ 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?

 

[jvicens]

Yes, your answer is correct. To see why, note that the summation is a geometric series with common ratio $\frac{1}{4}$, and since $x$ is in $l_{2}$, we know that $||x||^{2}=\sum_{k=1}^{\infty}|x_{k}|^{2}<\infty$. Therefore, as $n$ goes to infinity, the terms in the summation become smaller and approach 0, making the supremum equal to the first term in the series, which is $\frac{1}{4^{n+1}}$.