Let $A$ be a $\textrm{C}^*$-algebra and let $(a_n)$ be a sequence of elements of $A$. Is it then true that $$\sum_{n=1}^\infty a_n^*a_n\ \text{converges} \iff \sum_{n=1}^\infty a_na_n^*\ \text{converges?}$$
I've seen this stated as obvious, albeit in a more general setting in a paper, and I agree that it feels like something that might be true, but I don't seem to be able to get anywhere with it (other than noting its trivial if $A$ is abelian).
My only real attempt went along the lines of if $\sum_{n=1}^\infty a_n^*a_n$ converges, then it converges to some positive element $a^*a\in A$, then the obvious guess of what $\sum_{n=1}^\infty a_na_n^*$ might converge to is $aa^*$. But then I got no further.
It feels like quite an elementary question, but I'm completely stumped! Can anyone help with this?
In the algebra of all bounded operators on $\ell^2$, consider the operator $a_n$ defined by $$ a_n(\xi ) = \frac 1{\sqrt n}\langle \xi , e_1\rangle e_n, \quad\forall \xi \in \ell^2, $$ where $\{e_n\}_{n\geq 1}$ is the canonical basis. I'll leave it up to you to verify that $$\sum_{n=1}^\infty a_na_n^*$$ converges in norm, but $$\sum_{n=1}^\infty a_n^*a_n$$ doesn't converge, not even weakly!