By searching this url http://www2.math.ou.edu/~cremling/teaching/lecturenotes/fa-new/ln7.pdf in google given Theorem 7.4. and given its proff, but I do not understand very well this proof because it is not probably the best written proff think. I think that should be the case:
$(e-x)\sum_{n=0}^{\infty}x^n=(e-x)\lim_{N\to\infty}\sum_{n=0}^{N}x^n=\lim_{N\to\infty}(e-x)\sum_{n=0}^{N}x^n=\lim_{N\to\infty}\left(\sum_{n=0}^{N}x^n-\sum_{n=0}^{N}x^{n+1}\right)=\lim_{N\to\infty}(e-x^{N+1})=e$
1) Please tell me how should be written proff.
2) Please tell me why $\sum \Vert x^n \vert$ converges.
3) Tell me why $\sum_{n=0}^{\infty}x^n(e-x)=e$
Thanks for your answers.
In my opinion, there is not much that can be improved in the proof given but maybe some of the following will help.
We first have to note that $\sum_{n = 0}^\infty x^n$ is convergent. This can be done by calculating $\|\sum_{n = m}^N x^n \| \leq \sum_{n = m}^N \|x\|^n$ where $m \leq N$ and because $\sum_{n = 0}^\infty \|x\|^n$ is convergent (being a geometric series and $\|x\| < 1$) this shows that the sequence of partial sums $(\sum_{n = 0}^N x^n)_N$ is a Cauchy sequence, hence convergent (we are in a Banach space).
Now multiplication from any side is continuous and so we have $$ (e - x) \sum_{n = 0}^\infty x^n = \lim\limits_{N \rightarrow \infty} (e - x) \sum_{n = 0}^N x^n $$ and $$ \left(\sum_{n = 0}^\infty x^n\right) (e - x) = \lim\limits_{N \rightarrow \infty} \left( \sum_{n = 0}^N x^n (e - x) \right)$$ Now we calculate $(e - x) \sum_{n = 0}^N x^n = e - x^{N +1}$ and $ \sum_{n = 0}^N x^n (e - x) = e - x^{N +1}$ (telescoping sums) and take $N \rightarrow \infty$. Because $\|x\| < 1$ we have $x^{N +1} \rightarrow 0$ for $N \rightarrow \infty$, so all in all $$ (e - x) \sum_{n = 0}^\infty x^n = e = \sum_{n = 0}^\infty x^n (e - x).$$