Supremum calculation Calculate $\sup(\sum_{k=n+1}^{\infty}\frac{|x_{k}|^{2}}{4^{k}  })$, where $x=(x_{1},x_{2},....)$ is a member of $l_{2}$ and the supremum is take over all $x$ with $||x||= 1$.
My intuition says the answer is $\frac{1}{4^{n+1}}$. 

Your intuition is correct. It follows from that
$\sum_{k=n+1}^\infty\frac{|x_k|^2}{4^k} \leq \frac{1}{4^{n+1}}\sum_{k=n+1}^\infty|x_k|^2\leq \frac{1}{4^{n+1}}\sum_{k=1}^\infty|x_k|^2=\frac{1}{4^{n+1}}\|x\|_{l_2}^2=\frac{1}{4^{n+1}}$
thus the supremum is at most $\frac{1}{4^{n+1}}$. We also have that this value is obtained for $x\in l_2$ such that $x_k=1$ if $k= n+1$ and zero for all other indices. 
Therefore supremum of the sum must be equal to $\frac{1}{4^{n+1}}$.
