Suppose $H$ is a separable complex Hilbert space with inner product $(\cdot,\cdot)$ and norm $\|\cdot\|$, where $\|u\|^2 = (u,u)$. Suppose $u, u_1, u_2, \dots \in H$. Then $\lim_{n \to \infty} u_n = u$ strongly iff $\lim_{n \to \infty} \|u_n - u\| = 0$. $\lim_{n \to \infty} u_n = u$ weakly iff for all $v \in H$, $\lim_{n \to \infty} (u_n,v) = (u,v)$. Suppose $A, A_1, A_2 \dots$ are bounded operators on $H$. Then $\lim_{n \to \infty} A_n = A$ strongly iff for all $u \in H$, $\lim_{n \to \infty} A_n u = Au$ strongly. $\lim_{n \to \infty} A_n = A$ weakly iff for all $u, v \in H$, $\lim_{n \to \infty} (A_nu,v) = (Au,v)$, i.e., $\lim_{n \to \infty} A_n u = A_u$ weakly. (Note that convergence in the operator norm is a stronger notion that does not concern us here.) It is well known that weak operator convergence does not imply strong convergence, with the example given in the first answer below being standard. I had encountered a passage in which the author, having shown that $\lim_{n \to \infty} A_n = A$ weakly, concludes that $\lim_{n \to \infty} A_n = A$ strongly, giving by way of explanation the fact that---in the case at hand---$A, A_1, A_2, \dots$ are all unitary. This was not immediately obvious to me, but it is now; and I'll do penance by sketching the argument in case anyone is interested.
It is clearly enough to show this when $A$ is the identity operator; and for this it is enough to show that if $u, u_1, u_2, \dots$ are all unit vectors and $\lim_{n\to \infty} u_n = u$ weakly then $\lim_{n \to \infty} u_n = u$ strongly. To this end, suppose $\forall v \in H\,\,\lim_{n\to \infty} (u_n, v) = (u, v)$. Then, in particular, $\lim_{n\to \infty} (u_n, u) = (u, u) = 1$. Hence, $\lim_{n \to \infty} (u_n - u, u_n) = \lim_{n \to \infty} (u_n - u, u) = 0$, so $\lim_{n \to \infty} \|u_n - u\|^2 = \lim_{n \to \infty} (u_n - u, u_n - u) = 0$. Hence, $\lim_{n \to \infty} u_n = u$ strongly.