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Let $E$ be a Banach Algebra with identity, and $v\in E$, so that $||v|| < 1$. The geometric series $w = \sum_{k=0}^\infty v^k$ converges in the norm.

I can show that $||w|| \le \frac{1}{1-||v||}$, but does the equality hold? If it holds, how can I show this? If it doesn't hold, are there counterexamples?

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up vote 3 down vote accepted

$$\sum_{k=0}^\infty (-\tfrac{e}{2})^k = \left(\sum_{k=0}^\infty (-\tfrac{1}{2})^k\right)e = \frac{1}{1-\left(-\frac{1}{2}\right)}e = \tfrac{2}{3}e $$ because it's just a regular geometric series of real numbers times $e$, while $\frac{1}{1-\lVert -\frac{e}{2} \rVert} = 2$.

(Much belated) edit: corrected the summation.

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$$\sum_{k=0}^\infty (-\tfrac{e}{2})^k = \left(\sum_{k=0}^\infty (-\tfrac{1}{2})^k\right)e$$ I don't quite understand how they are equal. Shouldn't $$\left(\sum_{k=0}^\infty (-\tfrac{1}{2})^k\right)e = \sum_{k=0}^\infty (-\tfrac{e}{2^k})$$ Am I missing something? –  Mark Jul 17 '11 at 11:50
    
$e$ is the identity, so $e^k = e$ for all $k$ –  kahen Jul 17 '11 at 12:08
    
oh I see that makes sense now –  Mark Jul 17 '11 at 12:09
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You can. You have the following if $\|x\| < 1$: $$\sum_{n=0}^\infty \|x^n\| \le \sum_{n=0}^\infty \|x\|^n \le {1\over {1 - \|x\|}}.$$ This geometric sum is absolutely convergent and therefore convergent (Banach algebras are complete). It is not difficult to show that $$(e - x)^{-1} = \sum_{n=0}^\infty x^n,$$ where $e$ denotes the identity in the Banach algebra.

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Equality occurs on the positive real axis in the Banach algebra $\Bbb C$ inside of the open unit disk. –  ncmathsadist Jul 12 '11 at 22:22
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