Convergence of sequences in Hilbert spaces In a Hilbert space, let a sequence $(x_n)$ be weakly convergent to $x$ and be satisfied $\|x_{n+1}-x_n\|\to 0.$ I wonder if we can deduce that $(x_n)$ strongly converges to $x$, or at least it contains a subsequence strongly converges to $x$.
Thank you in advance for your help.
 A: This is not true. In $\ell^2$, you can consider the following sequence:
$$
e_1,
\frac12 e_2,
e_2,
\frac12 e_2,
\frac13 e_3,
\frac23 e_3,
e_3,
\frac23 e_3,
\frac13 e_3,
\frac14 e_4,
\frac24 e_4,
\frac34 e_4,
e_4,
\frac34 e_4,
\frac24 e_4,
\frac14 e_4,
\ldots$$
It is easy to check that this sequence has all desired properties.
Now, without a convergent subsequence:
$$
e_1, e_1 + \frac12 e_2,
e_1 + e_2,
\frac12 e_1 + e_2,
e_2,
e_2 + \frac13 e_3,
e_2 + \frac23 e_3,
e_2 + e_3,
\frac23 e_2 + e_3,
\frac13 e_2 + e_3,
e_3,
\ldots
$$
A: Edited heavily using In Hilbert space: $x_n → x$ if and only if $x_n \to x$ weakly and $\Vert x_n \Vert → \Vert x \Vert$. and the fact that weak convergence is unique:  If $$||x_{n+1}-x_n|| $$ converges to zero, then $(x_n)$ is Cauchy in H. So, $x_n$ converges to some $y$ in H. Using the other post, we have that $x_n$ converges weakly to $y$. But, weak convergence is unique. So, $x=y$ implying that $x_n$ converges to $x$ in H. Thus, $||x-x_n||$ tends to zero trivially giving us strong convergence.
