Explaining the theorem

Question

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 the third step, it looks like they've multiplied out $\lim(e-x)\sum_{n=0}^Nx^n$ but shouldn't the sum on the left be multiplied by $e$, i.e. $\lim\left(e\sum_{n=0}^Nx^n-\sum\ldots\right)$ ? — Jam, Aug 27 '14 at 13:04
@EulCan: I was confused at first, too, but it seems that $e$ is the unit element in the algebra, so its Ok. — Matthias Klupsch, Aug 27 '14 at 13:07
@user145717 Sorry, I'm not too familiar with the topic. $Matthias Ah, thanks for clarifying. — Jam, Aug 27 '14 at 13:08

score 1 · Accepted Answer · answered Aug 27 '14 at 13:40

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 $$ (e - x) \sum_{n = 0}^\infty x^n = e = \sum_{n = 0}^\infty x^n (e - x).$$

Explaining the theorem

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