Strong convergence to zero Let $(a^m)_{m\ge 1}$ be a sequence in a Hilbert space and also $$a^m=\sum_{n=1}^\infty a^m_n e_n$$ with $a^m_n\xrightarrow{m\rightarrow\infty}0,$ for any $n\ge 1.$
Does $a^m$ converges strongly to $0$ ?
 A: This is easily seen by noting that an inner product is defined by $<x,y>=\sum_{n=1}^{\infty}x_ny_n$. 
So for your question we can note $a^m=<a^m,e_n>\to<0,e_n>=0$ converges weakly, noting that $a^m\in H$ and $(e_n) $ an orthonormal sequence, for $n,m \in \mathbb{N}$.
Then to show that it isn't strongly convergent we let $n,p\in\mathbb{N}$ for $n\neq p$ then we have $||e_n-e_p||^2=<e_n,e_n>+<e_p,e_p>-<e_p,e_n>-<e_n,e_p> =2$, noting that $e_n,e_p$ are othogonal. 
Then if you need to show it converges weakly you should apply Bessel's inequality and apply the vanishing test.
A: I'm assuming that $(e_n)_n$ is an orthonormal set in $H$.
The answer is no. Consider $a^m = me_m$. We have $a^m_n = m\delta_{mn} \xrightarrow{m\to\infty} 0$ for all $n \in \mathbb{N}$. However, $(a^m)_m$ is not even a bounded sequence.

However, there is this result:

Let $(e_n)_n$ be an orthonormal basis for $H$. Let $(a_m)_m$ be a bounded sequence in $H$ such that there exists $a \in H$ with $\langle a_m, e_n\rangle \xrightarrow{m\to\infty} \langle a, e_n\rangle$ for all $n \in \mathbb{N}$. Then $(a_m)_m$ converges weakly to $a$.

