Do unitary transformations on Hilbert spaces preserve unconditional convergence? Let $H=(V,\langle \cdot,\cdot \rangle)$ be a separable complex Hilbert space with complex vector space $V$ and inner product $\langle \cdot,\cdot \rangle$. Since $H$ is separable, we can find a countable orthonormal basis $\{ e_k \}_{k \in \mathbb{N}}$ for $V$. Note that I do allow infinite dimensional vector space $V$.
Let $U \colon V \to V$ be a unitary transformation on $V$. Let $\psi = \sum_{k=1}^\infty \alpha_k e_k \in V$ with $\alpha_k \in \mathbb{C}$ and $\psi$ normalized: $\langle \psi,\psi \rangle = \sum_{k=1}^\infty \alpha_k \overline{\alpha_k}=1$.

Is the following computation correct?
$$ U\psi=U\left( \sum_{k=1}^\infty \alpha_k e_k \right) \stackrel{(1)}{=} \sum_{k=1}^\infty \alpha_k U e_k = \sum_{k=1}^\infty
 \alpha_k \sum_{l=1}^\infty \beta_{kl} e_l \stackrel{(2)}{=}
 \sum_{k=1,l=1}^\infty \alpha_k \beta_{kl} e_l \stackrel{(3)}{=} \sum_{l=1}^\infty
 \gamma_l e_l $$
  where $\gamma_l = \sum_{k=1}^\infty \alpha_k
 \beta_{kl}$

I am particularly worried about steps $(1)$, $(2)$ and $(3)$. Step $(1)$ seems to require some kind of linearity for infinity sums, and steps $(2),(3)$ seem to require that the sum converges unconditionally.
In case the computation is incorrect, are there reasonable additional assumptions that make it correct?
 A: The first step is a basic consequence of topology:
If $f: A \to B$ is a continuous function between two topological spaces then for every sequence $(x_n)_n$ of A convergent to an $x\in A$ you have that $(f(x_n))_n$ is convergent to a $f(x) $ in $B$ .
In the our case we have that $(\sum_{k=1}^{n}\alpha_ke_k)_n$ is a sequence convergent in $V$ to a 
$x=\sum_{k=1}^{\infty}\alpha_ke_k$ so, by continuity of $U$ you have that $U(\sum_{k=1}^{n}\alpha_ke_k)= \sum_{k=1}^{n}\alpha_kU(e_k) $ is convergent to $U(x)=U(\sum_{k=1}^{\infty}\alpha_ke_k)$.
So $U(\psi)= \sum_{k=1}^{\infty}\alpha_kU(e_k)$.
For the second and third case you can use this fact:
The scalar product is a continuous function with respect to topology induced by itself so if you want to calculate the $j^{th}$ component of $U(\psi) $ you have that:
\begin{align*}
\gamma_j=&\langle U\psi , e_j \rangle \\
=& \left\langle \lim_ {n\to\infty}\sum_{k=0}^{n}\alpha _kUe_k , e_j \right\rangle && \text{basic consequence of topology} \\
=& \lim_ {n\to\infty} \left\langle \sum_{k=0}^{n}\alpha _kUe_k , e_j \right\rangle && \langle \cdot , \cdot \rangle \text{ is continuous}\\
=& \lim_ {n\to\infty} \sum_{k=0}^{n}\alpha _k \langle Ue_k , e_j \rangle && \text{finite sum}\\
=& \lim_ {n\to\infty} \sum_{k=0}^{n}\alpha _k\beta_{kj} \\
=& \sum_{k=0}^{\infty}\alpha _k\beta_{kj}
\end{align*}
So you have that
$\sum_{k=0}^\infty\alpha_k\sum_{l=0}^{\infty}\beta_{kl}e_l=U\psi=\sum_{k=0}^{\infty}\gamma_ke_k$ 
and your second and third step is always justified. 
In any case you have dimmed that the identity is always true by the continuity of scalar product and property of a base of a vector space. 
