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Let $A_j$ be a sequence in $\mathbb {C}^{n\times n}$. Show that $\displaystyle \sum_{j=0}^\infty A_j$ converges if $\displaystyle \sum_{j=0}^\infty ||A_j||$ does.

Note that $\displaystyle \Vert A\Vert=\sup_{|x|=1} \vert Ax \vert$. I know that since $\displaystyle \sum_{j=0}^\infty ||A_j||$ does converge, then $\displaystyle \Vert A_j \Vert$ eventually converges to zero. Where do I go from here?

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I'm confused by your use of $j$. Sometimes it is a subscript, sometimes it seems like a vector or scalar multiple. – Emily Nov 4 '12 at 4:51
@ Ed sorry about that, I fixed it. – Klara Nov 4 '12 at 4:59
Notice that $\| \cdot \|$ as you give defines a norm and adapt the proof that an absolutely convergent series of real numbers converges. – Jose27 Nov 4 '12 at 6:22
up vote 1 down vote accepted

Hint: $\mathbb{C}^{n \times n}$ is complete, so we need only show that the sequence $S_n = \sum_{j=0}^{n} A_j$ is Cauchy.


Suppose without loss of generality that $n>m$. Then: $\|S_n-S_m\| =\|\sum_{j=m+1}^n A_j\| \leq \sum_{j=m+1}^n \|A_j\|= |\sum_{j=0}^n \|A_j\| - \sum_{j=0}^m \|A_j\||$ Since $\sum_{j=0}^\infty \|A_j\|$ is convergent, it is Cauchy, hence it follows that $S_n$ is also Cauchy.

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How'd you make the solution hidden like that? – littleO Nov 4 '12 at 9:21
Use >! at the start of the line (it is a bit fiddly, and I am sure there is a better way, but I am no 'La-techie'). Also, you can edit the answer to have a look (I think). – copper.hat Nov 4 '12 at 9:34
@copper you are genius. – Klara Nov 4 '12 at 13:14
@Klara: I am not, but your flattery is most appreciated :-). – copper.hat Nov 4 '12 at 15:46

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