Verifying a bound on the norm of an operator in $l_2$. The problem:

Define $L: l_2 \rightarrow l_2$ by $L(x_1, x_2, ...) = (y_1, y_2, ...)$, where $y_n = (x_1 + x_2 + ... + x_n)/n^2$. Show that $||L|| \leq (\sum_{n=1}^\infty 1/n^2)^{1/2}$.

My proof:

By the Cauchy-Schwarz inequality,$(\sum_{i=1}^n |a_i|)^2$ $= (\sum_{i=1}^n 1*|a_i|)^2$ $\leq(\sum_{i=1}^n 1^2)(\sum_{i=1}^n a_i) = n \sum_{i=1}^n |a_i|^2$ for any sequence of numbers $\lbrace a_i \rbrace$. Thus, for any x in $l_2$, $||Lx||^2$ $= \sum_{n=1}^\infty |(1/n^2) \sum_{i=1}^n x_i|^2$ $=\sum_{n=1}^\infty (|\sum_{i=1}^n x_i/n^2|)^2$ $\leq \sum_{n=1}^\infty (\sum_{i=1}^n |x_i/n^2|)^2$ $\leq \sum_{n=1}^\infty n \sum_{i=1}^n |x_i/n^2|^2$ $= \sum_{n=1}^\infty n \sum_{i=1}^n |x_i|^2/n^4$ $= \sum_{n=1}^\infty (n/n^4) \sum_{i=1}^n |x_i|^2$ $\leq \sum_{n=1}^\infty (1/n^3) ||x||^2$ $= ||x||^2 \sum_{n=1}^\infty 1/n^3$ $< ||x||^2 \sum_{n=1}^\infty 1/n^2$. Therefore, for any x in $l_2$, $||Lx|| < ||x||(\sum_{n=1}^\infty 1/n^2)^{1/2}$. In particular, $||L|| = sup \lbrace ||Lx||: ||x|| \leq 1 \rbrace \leq (\sum_{n=1}^\infty 1/n^2)^{1/2}$.

My concern is that my proof apparently gets a better bound than the problem statement suggests ($1/n^3$ instead of just $1/n^2$), which makes me suspicious that I'm missing a "*n" somewhere, but even after looking carefully at it a few times, I can't find anything. Any help verifying or finding an error would be appreciated.

Right. We have
$$\left\lvert \sum_{k=1}^n x_k\right\rvert\leqslant \sum_{k=1}^n \lvert x_k\rvert \leqslant \sqrt{n}\cdot \lVert x\rVert_{\ell^2}$$
by Cauchy-Schwarz. So if $y_n = \frac{1}{n^\alpha}\sum_{k=1}^n x_k$, we have
$$\lvert y_n\rvert \leqslant \frac{1}{n^{\alpha-1/2}}\lVert x\rVert_{\ell^2},$$
and hence
$$\lVert L_\alpha(x)\rVert_{\ell^2}^2 \leqslant \lVert x\rVert_{\ell^2}^2\sum_{n=1}^\infty \frac{1}{n^{2\alpha-1}}$$
for all $x\in \ell^2$, or
$$\lVert L_\alpha\rVert \leqslant \left(\sum_{n=1}^\infty \frac{1}{n^{2\alpha-1}}\right)^{1/2}$$
for all $\alpha > 1$. Taking $\alpha = 2$ gives your bound $\sqrt{\zeta(3)}$, which indeed and correctly is smaller than the given $\sqrt{\zeta(2)} = \frac{\pi}{\sqrt{6}}$.
