convergence problem in space of sequences

Question

I want to prove that if a sequence is Cauchy in $l^2(\Bbb N)$, then it converges in $l^2(\Bbb N)$. I proved the convergence part, but I've been in trouble with $a\in l^2$ s.t for Cauchy sequence $a_n\in l^2(\Bbb N)$, $a_n\to a$. I got $$\sum_{i=1}^n|a(i)|^2\le \sum_{i=1}^n|a_n(i)|^2+\sum_{i=1}^n|a_n(i)-a(i)|^2+2\sum_{i=1}^n|a_n(i)||a_n(i)-a(i)|$$and as $n$ goes to infinity, I know that $\sum_{i=1}^n|a_n(i)|^2,\ \sum_{i=1}^n|a_n(i)-a(i)|^2$ are bounded, but I can't tell so is $2\sum_{i=1}^n|a_n(i)||a_n(i)-a(i)|$.

Answer 1

Write $|a(i)|^2\leqslant 2\cdot |a(i)-a_n(i)|^2+2|a_n(i)|^2$, and summing, $$\tag{1}\sum_{i=1}^N|a(i)|^2\leqslant 2\cdot \sum_{i=1}^N|a(i)-a_n(i)|^2+2\sum_{i=1}^N|a_n(i)|^2$$ (note that we take a capital $N$, hence not the same as the index). Notice that a Cauchy sequence is bounded, hence there is a constant $C$ such that for each $n$, $\sum_{i=1}^\infty|a_n(i)|^2\leqslant C$. Plugging this in (1), we get for each $n$, $$\sum_{i=1}^N|a(i)|^2\leqslant 2\cdot \sum_{i=1}^N|a(i)-a_n(i)|^2+2C.$$ Then take the limit $n\to \infty$: $\sum_{i=1}^N|a(i)|^2\leqslant 2C$ and we are done since $N$ is arbitrary.