Geometric Sums in Banach Algebra

Question

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?

Answer 1

$$\sum_{k=0}^\infty (-\tfrac{e}{2})^k = \left(\sum_{k=0}^\infty (-\tfrac{1}{2})^k\right)e = \frac{1}{1-\left(-\frac{1}{2}\right)}e = \tfrac{2}{3}e $$ 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.

Answer 2

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.