$\lim_{n}||T_n||\not\to 0\implies\sum_{n=0}^\infty T_n$ cannot converge in operator norm topology for self-adjoint $T_n\in\mathcal{B}(\mathcal{H})$

Question

Let $\mathcal{H}$ be a Hilbert space and $\{T_n\}$ be a sequence of self-adjoint bounded linear operators. Take $T:=\sum_{n=0}^\infty T_n$ for now just as an assignment without worrying whether the series on the RHS converges (in anything). I am reading a book which effectively states that if the norms of $T_n$s do not converge to zero as $n\to\infty$, then $\sum_{n=0}^\infty T_n$ cannot converge in operator norm topology. I know how to prove that $\sum_{n=0}^\infty ||T_n|| < \infty \implies \sum_{n=0}^\infty T_n\in\mathcal{B}(\mathcal{H})$, but I am not aware of this converse implication and it might even be wrong for a general bounded linear operator. But is this true for a self-adjoint operator?

Answer 1

In any normed space, if you have a sequence of vectors $x_{n}$ such that $\sum_{n=1}^{\infty}x_{n}$ converges then $||x_{n}||\to 0$.

This is simply because the sequence of partial sums $s_{n}=\sum_{k=1}^{n}x_{k}$ is then convergent and hence Cauchy.

In particular given $||s_{n}-s_{m}||<\epsilon$ for all $m,n\geq N$. and so in particular $||s_{n+1}-s_{n}||=||x_{n+1}||<\epsilon\,,\forall n\geq N$ .

So the contrapositive must also be true. That is , if $||x_{n}||$ does not go to $0$ then $\sum_{k=1}^{\infty}x_{k}$ cannot converge.