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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.

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  • $\begingroup$ The calculus $\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} $ is perfect, you must isolate $x$ and note that $S_{i+1} - S_i = a_{i+1} $ for all $i.$ $\endgroup$
    – Ilovemath
    Jun 3, 2020 at 16:07
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    $\begingroup$ You were correct to doubt this as written. Manipulations of series written as $a_1 + a_2 + \ldots $ where convergence has not been established can be misleading. For example, $1 - 1 + 1- 1+ \ldots = 1 -(1-1) -(1-1) - \ldots = 1$. In this case the sum in question is convergent -- not necessarily absolutely convergent. $\endgroup$
    – RRL
    Jun 3, 2020 at 18:09

1 Answer 1

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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

$$(1-x)\sum_{n=0}^{\infty}s_{n+1} x^n = \lim_{m \to \infty} (1-x)\sum_{n=0}^{m-1}s_{n+1} x^n = \lim_{m \to \infty}\sum_{n=0}^ma_n x^n- \lim_{m \to \infty}s_{m+1}x^m \\ = \sum_{n=0}^\infty a_n x^n$$

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    $\begingroup$ Since |x| < 1 we have $\lim_{n \to \infty} x^n = 0$. $\endgroup$
    – RRL
    Jun 3, 2020 at 16:06
  • $\begingroup$ Thank you very much, this makes complete sense to me. Would it be correct to say the way it is presented is at least 'informal' in some sense? In particular, when they do the 'rearrangement', can we really group all positive and negative terms in this way? What you've shown makes perfect sense to me, but the steps in the proof seem fundamentally different to the way you've shown. In particular the grouping of terms is bugging me because I don't think such a bijection exists? $\endgroup$
    – masiewpao
    Jun 4, 2020 at 10:14
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    $\begingroup$ @masiewpao: You're welcome. As these are power series, we have absolute convergence within the radius of convergence so such rearrangement is possible. I prefer when possible to operate on the finite sum (where rearrangement is always valid) and show the remainder converges to $0$. $\endgroup$
    – RRL
    Jun 4, 2020 at 13:50
  • $\begingroup$ I think my confusion is with whether or not $s_1 + s_2x + s_3 x^2 + s_4 x^3 + ... + s_{n+1}x^n + ... - s_1 x - s_2 x^2 - s_3 x^3 - ... -s_n x^n - s_{n+1} x^{n+1}-...$ is actually a rearrangement of the original series. What I mean is that I'm not sure if the above series is actually a rearrangement of the original series such that we can conclude absolute convergence. For example with the way it's written, there is no $n\in \mathbb{N}$ such that the nth term of the 'rearranged' series is any negative term, whereas there exists $n$ in the original series such that the nth term is negative. $\endgroup$
    – masiewpao
    Jun 4, 2020 at 14:17
  • $\begingroup$ Although I also realise that I am being a bit pedantic, as I think this is just a notational issue because what you've done with the limits made ite very explicit for me what is actually going on! $\endgroup$
    – masiewpao
    Jun 4, 2020 at 14:19

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