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Suppose $A_k>0$ (which means they are positive definitive square $n\times n$-matrices with $n>1$). If $\sum_{k=1}^\infty A_k$ exists, then $\sum_{k=1}^\infty \|A_k\| < +\infty$, Where $\|A\|=\sup_{\|x\|\leq 1}\langle Ax,x\rangle$.

Is this true? (I am not able to give any counterexample.)

Thank you!

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    $\begingroup$ In $\sum_{k=1}^\infty A_k = \lim_{K\to\infty} \sum_{k=1}^K A_k$, is the limit taken elementwise? $\endgroup$
    – Jochen Voss
    Commented Nov 3, 2016 at 10:26

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You can bound $\|A_k\| \leq C(n)\max_{i,j} |(A_k)_{ij}|$ for some function of the dimension only $C(n)$, because all norms are equivalent in finite dimension. If I am not mistaken $C(n)=\sqrt{n}$, but it doesn't really matter here.

This maximum is attained on a (positive) diagonal entry, because of positive definiteness.

Then you have $$\sum_k \|A_k\| \leq C(n) \sum_k \max_i (A_k)_{ii} \leq C(n)\sum_k \sum_i (A_k)_{ii} = C(n) \sum_i (\sum_k (A_k)_{ii}),$$ which is finite because $\sum_k (A_k)_{ii}$ is finite for each $i$.

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Yes. The norm of a positive definite matrix does not exceed its trace, and the sum of traces is finite, since the sum of diagonal elements is finite for each of $n$ places.

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    $\begingroup$ Neat - this is simpler and more elegant than my solution; it deserves more upvotes. $\endgroup$
    – Federico Poloni
    Commented Nov 3, 2016 at 14:33

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