Suppose $a,b \in l^2(\mathbb{N})$. Show that there exists $C$ such that $$\sum_{n=1}^{\infty} \sum_{k=1}^\infty \frac{a_n b_k}{n+k} \leq C\ ||a||\ ||b|| $$
My (incorrect) attempt: I thought of using a double cauchy-schwarz argument. We can factor the $a_n$ out of the inner sum (duh):
$$\sum_{n=1}^{\infty} \sum_{k=1}^\infty \frac{a_n b_k}{n+k} = \sum_{n=1}^{\infty} a_n \sum_{k=1}^\infty \frac{b_k}{n+k} $$
Now applying Cauchy-Schwartz to the inner sum first:
$$\sum_{k=1}^{\infty} \frac{b_k}{n+k} \leq \bigg( \sum_k |b_k|^2 \bigg)^{1/2} \bigg(\sum_k \frac{1}{(n+k)^2} \bigg)^{1/2} = ||b|| \bigg(\sum_{k=1}^{\infty} \frac{1}{(n+k)^2} \bigg)^{1/2} = c_n$$
Then we can apply it again:
$$\sum_{n=1}^{\infty} a_n \sum_{k=1}^\infty \frac{b_k}{n+k} \leq \sum_{n=1}^{\infty} a_nc_n \leq \bigg( \sum_n |a_n|^2 \bigg)^{1/2} \bigg( \sum_n |c_n|^2 \bigg)^{1/2}$$
$$\leq ||a||\ ||b|| \bigg( \sum_{n=1}^{\infty} \sum_{k=1}^\infty \frac{1}{(n+k)^2}\bigg)^{1/2} $$
The problem is that the double sum in the last line does not converge. In fact,
$$ \sum_{n=1}^{\infty} \sum_{k=1}^\infty \frac{1}{(n+k)^2} = \sum_{n=1}^{\infty} \frac{n}{(n+1)^2} $$
Does anyone have any other ideas?