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I have a question about the convergence of the Neumann series:

Let $A$ be a matrix with spectral radius $\rho(A)<1$, i.e., all eigenvalues of $A$ are strictly less than $1$. Does that imply that the series \begin{equation} \sum_{i=0}^{\infty}A^i \end{equation} converges (in the operator norm)? I know how to prove it if the operator norm of $A$ is strictly less than $1$, but I don't know how to prove it if I only know that the spectral radius is less than $1$.

Many thanks for any help!

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    Are you familiar with Gelfand's formula relating operator norm to spectral radius?2012-11-22
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    I know the formula, but that's it. in particular, I would not know how to apply it here...2012-11-22
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    If $\|A^n\|^{1/n} \to c < 1$ then for some $n$ large enough, $\|A^n\| < 1$.2012-11-22
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    ok, I understand. but how does that help me here exactly. sorry that I don't see it... and many thanks for your help!2012-11-22
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    You said you know how to show convergence given the operator norm is $<1$...How does the proof go?2012-11-22
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    $\|A^n\|^{1/n}\leq d<1$ implies $\|A^n\|\leq d^n$, so you can use convergence of geometric series.2012-11-22
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    well, how to use your hint in the proof with the operator norm smaller than 1 is what I was thinking about. But I cant directly use sub-multiplicativity of the operator norm to deduce from your hint that the operator norm is less than 1. the key element in the proof is to show that the partial sums are Cauchy. how could I deduce the same property here? I can't come up with a geometric series as majorant (or do I - am I just not able to see it?)?2012-11-22
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    ok, there is the answer - I was just not able to see it! sorry for taking your time. and many thanks Erick and Jonas!!! I really appreciate your help.2012-11-22

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