Why weak convergence doesn't imply convergence? Let $(H,\langle,\rangle)$ be a Hilbert space. We have that $u_n$ converges to $u$ weakly if
$$\lim_{n\to \infty}\langle u_n,v\rangle=\langle u,v\rangle$$
for all $v\in H$. But why doesn't it converge strongly? Indeed, if $v=u_m$, then,
$$\lim_{n\to \infty }\langle u_n,u_m\rangle=\langle u,u_m\rangle$$
for all $m$, and thus
$$\lim_{m\to \infty }\lim_{n\to \infty }\langle u_n,u_m\rangle=\langle u,u_m\rangle\lim_{m\to \infty }\langle u,u_m\rangle=\langle u,u\rangle$$
therefore,
$$\lim_{n\to \infty }\langle u_n,u_n\rangle=\langle u,u\rangle\implies \lim_{n\to \infty }\|u_n\|=\|u\|$$
and thus it converges weakly. What's wrong here?
 A: You have used
$$\lim_{m \to \infty} \lim_{n \to \infty} a_{n,m} = a$$
implies
$$\lim_{n \to \infty} a_{n,n} = a.$$
However, this is not true.
Consider, e.g.,
$$a_{n,m} = \begin{cases} \pi & m < n, \\ \mathrm{e} &n < m, \\ 42 & m = n.\end{cases}$$
Edit: It is also enlightening to consider the most famous weakly convergent sequence. Let $\{u_n\}_{n \in \mathbb N}$ be an orthonormal system. Then, your $a_{n,m}$ satisfies
$$a_{n,m} = \begin{cases} 0 & m \ne n, \\ 1 &n = m.\end{cases}$$
Again,
$$\lim_{m \to \infty} \lim_{n \to \infty} a_{n,m} = \lim_{n \to \infty} a_{n,n}$$
fails blatantly.
A: This can be understood geometrically.
To makes things easier, specify the limit point to be zero: $u = 0$. The general case $u \neq 0$ follows from affine translation.
The set 
$$\{ u : \langle u, v \rangle \le \epsilon \}$$
describes a semi-infinite slab contained between two closely spaced parallel hyperplanes with normal vector $v$. The convergence
$$\langle u_n, v \rangle \rightarrow 0$$
occurs successfully if for every slab width, nomatter how thin, a tail of the sequence is contained within the slab. I.e., the sequence is asymptotically "controlled" in the $v$-direction. The sequence converges weakly if this holds for every possible slab direction. I.e.,
for each direction v:
    for each width ε:
        A tail of the sequence is contained in the slab with 
        direction v and width ε.

In contrast, the sequence converges strongly if every scaled copy of the unit ball,
$$\{ u : ||u|| \le \epsilon \},$$
nomatter how small, contains a tail of the sequence:
for each width ε:
    A tail of the sequence is contained within the ball of radius ε.

Strong convergence controls the convergence in all directions simultaneously.
In infinite dimensions a problem occurs since one can create a well-spaced infinite sequence without going anywhere, by cycling through the infinity of different dimensions:
\begin{align*}
u_1 &= (1,0,0,\dots) \\
u_2 &= (0,1,0,\dots) \\
u_3 &= (0,0,1,\dots) \\
\vdots &
\end{align*}
The sequence converges weakly since every direction is controlled eventually, but it does not converge strongly since you have to wait infinitely long for every direction to be controlled.
