# Show operator is compact

I'm currently studying for a functional analysis exam and I noticed that often when it is asked to prove that an operator is compact I run into issues:

Let H be a real Hilbert space. Use the theorem of Banach-Alaoglu to show the following operator is compact:

$$Tx:=\sum_{k=1}^{\infty}y_n$$ where $$x_n$$ and $$y_n$$ are sequences in H with $$\sum_{k=1}^{\infty}||y_n||*||x_n||<\infty$$.

I know that for a bounded sequence $$(x_n)_{n \in \mathbb{N}}$$ I need to show that that $$(Tx_n)_{n \in \mathbb{N}}$$ has a strongly convergent subsequence. B-A provides me a weakly convergent subsequence of $$(x_n)_{n \in \mathbb{N}}$$ and a weakly convergent subsequence of $$(Tx_n)_{n \in \mathbb{N}}$$ because via Cauchy-Schwarz it follows that T is bounded as well. How do I continue from here? While I'm foremost interested in knowing how to solve this via B-A I'd be interested in hearing other approaches that might come in handy for this type of exercise as well.

A different approach: Let $$T_N(x)= \sum\limits_{n=1}^{N} \langle x, x_n \rangle y_n$$. Then,$$T_n$$ has finite rank and all finite rank operators are compact. Also, $$\|T-T_N\|\leq \sum\limits_{n=N+1}^{\infty} ||x_n\|\|y_n\|\to 0$$ as $$N \to\infty$$. This implies that $$T$$ is compact.