I'm trying to prove the following statement with $E = \ell^2$ with usual norm $||x||_{\ell^2}$ and complex coefficients. Define $T:E \to E$ such that $$ u = (u_n)_{n\in\mathbb{N}} \mapsto T(u) = \left(\sum_{k=1}^\infty \frac{u_k}{n+k}\right)_{n\in\mathbb{N}} $$ show that $T$ isn't a compact operator.
My attepm: Define a sequence $(w_n^m)_{n\in\mathbb{N}}$ such that $w_n^m$ weakly converge to $0$ but $T(w_n)$ diverge. For example, let $$ w^m = \frac{1}{\sqrt{n}}e_n = \frac{1}{\sqrt{n}}(1,\ldots,1,0,\ldots). $$ the sequence is one until the $m$ position. Actually, $w_n$ weakly converge to $0$ but I can't prove that the image diverge. Any suggestion? Also, if anyone could suggest bibliography about this kind of operators I would be appreciated
UPDATE: I correct the statement of the excercise.
UPDATE 2: Note that
\begin{align} T(w^m) & = \left(\sum_{k=1}^\infty \frac{w_n^m}{k+n}\right)_{n\in\mathbb{N}} \\ & = \left(\sum_{k=1}^m \frac{1}{\sqrt{m}}\cdot\frac{1}{k+n}\right)_{n\in\mathbb{N}}\\ & = \left(\frac{1}{\sqrt{m}} (\psi(m+n+1)-\psi(n+1)) \right)_{n\in\mathbb{N}} \\ \end{align}
where $\psi$ is the digamma function. Can I argue that for each $n$ $$ \frac{1}{\sqrt{m}} (\psi(m+n+1)-\psi(n+1)) \to \infty $$ I'm not sure about properties of digamma function.
Note that
$$ \sum_{j=1}^m \frac{1}{j+n} \geq \int_i^{m+1} \frac{dt}{t+n} \geq \log\left(1+\frac{m}{n+1}\right) $$ with this
$$ ||Tw^m||^2 \geq \left(\frac{m-1}{m}\right) \log^2(2) $$ when $m\to \infty$ following the desire result.