Completeness of $\ell^2$ and equivalence of weak and strong convergence in $\ell^1$ I'm looking at two different functional analysis exercises, and I feel that I've hit the same road block with them. I was wondering if there is a kind of "systematic" way of fixing these solutions (or if I need to abandon hope and try a new approach).
The first exercise asks to show that $\ell^2$ is complete. My approach is

Take $\{f_n\}$ Cauchy in $\ell^2$ and observe that $\{ f_n(m) \}$ is Cauchy in $\mathbb{C}$ for each natural number $m$. I define $f \in \ell^2$ by $f(m) := \lim_{n \to \infty} f_n(m)$, but when I try to show that $f_n \rightarrow f$, I end up looking at the sum $$\sum_{j=1}^\infty |f_n(j)-f(j)|^2,$$ and trying show that it tends to $0$. I can control each individual summand, but since I have to deal with infinitely many, I'm not sure how to proceed.

The second exercise asks to show that weak convergence implies strong convergence in $\ell^1$. My approach is

Take $\{f_n\}$ weakly convergent in $\ell^1$. Then $$|| f_n - f || = \sum_{j=1}^{\infty} |f_n(j)-f(j)| = \sum_{j=1}^{\infty} |\pi_j(f_n-f)|,$$
  where $\pi_j$ is the projection to the $j$-th coordinate of $f$. Again, I can control each of these summands (by weak convergence and continuity of these projections), but I don't know how to deal with infinitely many. (For this particular exercise, the hint is to use the fact that $\ell^1$ is dual to $\ell^{\infty}$, but I'm still trying to use this).

To reiterate: I'd like to know if there's some general technique for dealing with these situations. If so, what do I do, and how should I think about it? If not, are there any possible ad hoc fixes, or should I pursue a different course of action? 
Thanks in advance!
 A: For the first question, since $\{f_n\}$ is a Cauchy sequence, for each $\varepsilon>0$ there is an $N$ such that if $n,m\geq N$ then
$$ \sum_{j=1}^{\infty}|f_n(j)-f_m(j)|^2<\varepsilon^2 $$
Let $f(j)=\lim_{n\to\infty}f_n(j)$ as in your question. Then for each fixed $J$, we have
$$ \sum_{j=1}^{J}|f_n(j)-f_m(j)|^2\leq \sum_{j=1}^{\infty}|f_n(j)-f_m(j)|^2<\varepsilon^2 $$
if $m,n\geq N$, hence taking $m\to\infty$ in the first sum yields
$$ \sum_{j=1}^{J}|f_n(j)-f(j)|^2<\varepsilon^2$$
Then taking $J\to\infty$ shows that
$$ \sum_{j=1}^{\infty}|f_n(j)-f(j)|^2\leq \varepsilon^2$$
hence $f_n\to f$ in $\ell^2$.
A: First, a Cauchy sequence is always bounded. This is because there exists $N$ such that $\|f_n-f_k\| < 1$ whenever $n,k \ge N$. Then
$$
            \|f_k\| \le M=\max\{\|f_1\|,\|f_2\|,\cdots,\|f_N\|+1\},\;\;\; k=1,2,3,\cdots.
$$
Another standard thing is to note that the point evaluations are continuous:
$$
       |f_j(n)-f_{k}(n)| \le \|f_j-f_k\|,
$$
which means that $f(n)=\lim_jf_j(n)$ exists for all $n$ because the field $\mathbb{C}$ or $\mathbb{R}$ is complete. So, for any fixed upper limit $N$,
$$
          \sum_{n=1}^{N}|f_j(n)|^2 \le \|f_j\|^2 \le M^2.
$$
The above holds for all $j$ and for all $N$. Hence, the following holds for all $N$:
$$
     \sum_{n=1}^{N}|f(n)|^2 \le \lim_{j}\sum_{n=1}^{N}|f_j(n)|^2 \le M^2.
$$
Therefore $f\in \ell^2$. For every $\epsilon > 0$ there exists $J$ such that $\|f_j-f_k\| < \epsilon/2$ for all $j,k \ge J$. Therefore, for all $N$,
$$
    \sum_{n=1}^{N}|f_j(n)-f_k(n)|^2  < \frac{\epsilon^2}{4},\;\;\; j,k \ge J \\
    \sum_{n=1}^{N}|f(n)-f_k(n)|^2 \le \frac{\epsilon^2}{4},\;\;\; k \ge J.
$$
Because this holds for all $N$,
$$
         \sum_{n=1}^{\infty}|f(n)-f_k(n)|^2 \le \frac{\epsilon^2}{4} < \epsilon^2,\;\;\; k \ge J.
$$
Because $\epsilon > 0$ was arbitrary, by the definition of the vector limit,
$$
             \lim_{j} f_j = f.
$$
