In my book a proof of Abel's Limit Theorem is presented. The theorem as stated: Let$f(x)$ be the sum function of the power series $\sum_{n=0}^\infty a_nx^n$, which has radius of convergence 1; and let $\sum_{n=0}^\infty a_n$ be convergent. Then $\lim_{x \to 1^-} f(x) = \sum_{n=0}^\infty a_n$.
I am having trouble at the very start of the proof where we manipulate some series. We start by defining $s_n = a_0 + a_1 + a_2 +…+ a_{n-1}$ to be the nth partial sum of $\sum_{n=0}^\infty a_n$. We let $s$ denote the sum $\sum_{n=0}^\infty a_n$. We also note then that $a_0 = s1$ and $a_n = s_{n+1} – s_n$ for all $n \geq1$.
The first part of the proof (immediately following the prior definitions):
It follows that for $\mid x \mid < 1$,
\begin{align}
(1-x) \sum_{n=0}^\infty s_{n+1}x^n & = (1-x)(s_1 + s_2 x + s_3 x^2 + …+s_{n+1} x^n + …)\\
& = s_1 + s_2x + s_3 x^2 + s_4 x^3 + … + s_{n+1}x^n +\\
&-s_1 x – s_2 x^2 – s_3 x^3 – … -s_n x^n – s_{n+1} x^{n+1}-…\\
& =s_1 + (s_2-s_1)x +(s_3-s_2)x^2 + … +(s_{n+1} – s_n)x^n + …\\
& = \sum_{n=0}^\infty a_n x^n = f(x)
\end{align}
(Just as a quick note I have presented the equations exactly as presented in the book, after the $s_{n+1}$ term on the second line, there is no "$…$" after the $+$ symbol; I just assumed this was a small error but included it in case.)
I don't understand how we go from the RHS of line 1, to the infinite series (over lines 2 and 3). I can't figure out how to rigorously justify this manipulation. As a first step, I am not even sure how we know that $\sum_{n=0}^\infty s_{n+1}x^n$ is even convergent. But even if I assume it is absolutely convergent, I still can't work out how to justify this step:
I know that $$(1-x)(s_1 + s_2 x + s_3 x^2 + …+s_{n+1} x^n + …)$$ is equal to
$$(s_1 – s_1x) + (s_2x-s_2x^2) + (s_3x^2 – s_3x^3)+…+(s_{n+1}x^n – x^{n+1})+…$$ because I have just assumed the series is absolutely convergent for the time being, so I can multiply the constant in without changing the value of the series. However I am not really sure about going from this to the the third line: $$s_1 + s_2x + s_3 x^2 + s_4 x^3 + … + s_{n+1}x^n + … – s_1 x – s_2 x^2 – s_3 x^3 – … -s_n x^n – s_{n+1} x^{n+1}-…$$
Firstly, can we simply 'unbracket' the terms in $$(s_1 – s_1x) + (s_2x-s_2x^2) + (s_3x^2 – s_3x^3)+…+(s_{n+1}x^n – x^{n+1})+…$$
I.e. can we say that $$(s_1 – s_1x) + (s_2x-s_2x^2) + (s_3x^2 – s_3x^3)+…+(s_{n+1}x^n – x^{n+1})+… = s_1 – s_1x + s_2x-s_2x^2 + s_3x^2 – s_3x^3+…+s_{n+1}x^n – x^{n+1}+…$$
Secondly, assuming we can unbracket the terms, all the positive terms are grouped at the start, and then followed by all the negative terms after. The reason I am confused is because it seems as if the partial sums of this rearrangement will be increasing, and never take into account any negative terms. Meaning that $s_n$ for any $n \in \mathbb{N}$ will be positive, and so the negative terms are never 'reached'. Therefore I wouldn't expect the series $$(s_1 – s_1x) + (s_2x-s_2x^2) + (s_3x^2 – s_3x^3)+…+(s_{n+1}x^n – x^{n+1})+…$$ and $$s_1 + s_2x + s_3 x^2 + s_4 x^3 + … + s_{n+1}x^n + … – s_1 x – s_2 x^2 – s_3 x^3 – … -s_n x^n – s_{n+1} x^{n+1}-…$$ to have the same value.
At the same time if I assume that the series is absolutely convergent, then I know there is a theorem that tells us that any rearrangement is also absolutely convergent. I am not sure how to square away this apparent contradiction. I suspect that this is because this grouping of the terms is not actually valid since it is not a bijection from the original series to the 'grouped' series. But I don't then know how to justify the step in the proof.
To summarise:
(1) Must we know that $\sum_{n=0}^\infty s_{n+1}x^n$ is convergent to perform the steps of this proof?
(2) Even assuming the absolute convergence of $\sum_{n=0}^\infty s_{n+1}x^n$, what is the justification we can use to say the series $(s_1 – s_1x) + (s_2x-s_2x^2) + (s_3x^2 – s_3x^3)+…+(s_{n+1}x^n – x^{n+1})+…$ is equal to the series $s_1 + s_2x + s_3 x^2 + s_4 x^3 + … + s_{n+1}x^n + … – s_1 x – s_2 x^2 – s_3 x^3 – … -s_n x^n – s_{n+1} x^{n+1}-…$? Why can we unbracket and rearrange? In particular, it doesn't seem to me like this is an actual rearrangement because I don't think the bijection between the series exists.
Thanks very much.
Best Answer
The manipulations are based (tacitly) on the assumption that $\sum s_{n+1} x^n$ converges. However, this can be proved in the process of deriving your first equation.
The argument is in effect using summation by parts. More clearly, with $s_0 = 0$, we have
$$\tag{*}\begin{align}\sum_{n=0}^ma_n x^n &= \sum_{n=0}^m(s_{n+1}-s_n) x^n \\&= s_{m+1} x^m + \sum_{n=0}^{m-1}s_{n+1} x^n - \sum_{n=0}^{m}s_{n} x^n \\ &= s_{m+1} x^m + \sum_{n=0}^{m-1}s_{n+1} x^n - s_0 -\sum_{n=1}^{m}s_{n} x^n \\ &= s_{m+1}x^m+ \sum_{n=0}^{m-1}s_{n+1} x^n -\sum_{n=0}^{m-1}s_{n+1} x^{n+1} \\ &= s_{m+1}x^m+ \sum_{n=0}^{m-1}s_{n+1} (x^n - x^{n+1})\\ &= s_{m+1}x^m+ (1-x)\sum_{n=0}^{m-1}s_{n+1} x^n\\ \end{align}$$
Note that $s_{m+1} \to \sum_{n=0}^\infty a_nx^n$ and $x^m \to 0$ as $m \to \infty$. Rearranging (*) and taking limits, the sum on the RHS converges with
$$(1-x)\sum_{n=0}^{\infty}s_{n+1} x^n = \lim_{m \to \infty} (1-x)\sum_{n=0}^{m-1}s_{n+1} x^n = \lim_{m \to \infty}\sum_{n=0}^ma_n x^n- \lim_{m \to \infty}s_{m+1}x^m \\ = \sum_{n=0}^\infty a_n x^n$$