a) parallelepiped {$x \in \ell^2, |x_n| \leq a_n$}
b) ellipsoid {$x\in \ell^2, \sum_{n=1}^{\infty} \frac{|x_n|^2}{|a_n|^2} \leq 1$
I've already shown that a works if $|a_n| \leq \frac{1}{n}$, but I really don't know what to do here because thats definitely not all I need
ANSWER. A necessary and sufficient condition for compactness, in both cases, is that $\,a_n\to 0$.
HINTS. First all all, if $x^k=(x^k_n)_{n\in\mathbb N}$, $k\in\mathbb N$, using the fact that $|x_n^k|$, is bounded in $k$, for all $n$, and a diagonal argument one can pick a subsequence of $(x^k)$, cal it $(y^k)$, such that $y_n^k\to y_n$, for all $n$. Of course, this DOES NOT imply that $y^k\to y$ in the $\ell^2-$sense. It converges weakly.
In fact, if $y^k\to y$ weakly, and $|y^k|_{\ell^2}\to |y|_{\ell^2}$, then $y^k\to y$ strongly - In general, if $y^k\to y$ weakly, then $\limsup_{k\to\infty} |y^k|_{\ell^2}\le |y|_{\ell^2}$.
It remains to show that, $a_n\to 0$, if and only if $|y^k|_{\ell^2}\to |y|_{\ell^2}$.