# Which is the Abel's theorem invoked in the context of convergence of this infinite product?

Motivation: As I wrote in this answer the following product is evaluated in Proposition 5 of Jean-Paul Allouche and Jeffrey Shallit's paper The ubiquitous Prouhet-Thue-Morse sequence

$$P=\displaystyle\prod_{n=0}^{\infty }\left( \frac{2n+1}{2n+2}\right) ^{(-1)^{t_{n}}},$$

where $\left( t_{n}\right) _{n\geq 0}=\left( 0,1,1,0,1,0,0,1,\ldots \right)$ is a binary sequence defined recursively by $t_{0}=0,t_{2n}=t_{n}$ and $% t_{2n+1}=1-t_{n}$ for $n\geq 0$. According to the authors the product is "convergent by Abel's theorem", but I have no idea which theorem is this. Due to the properties of $\left( t_{n}\right)$ the terms of $P$ can be broken into two groups of two terms each (this idea came from Ross Millikan). I got:

$$\begin{eqnarray*} P &=&\displaystyle\prod_{m=0}^{\infty }\left( \frac{8m+1}{8m+2}\frac{8m+4}{8m+3}\cdot \frac{8m+6}{8m+5}\frac{8m+7}{8m+8}\right) ^{(-1)^{t_{m}}} \\ &=&\displaystyle\prod_{m=0}^{\infty }\left( 1-\frac{1}{32m^{2}+20m+3}\right) ^{(-1)^{t_{m}}}\displaystyle\prod_{m=0}^{\infty }\left( 1+\frac{1}{32m^{2}+52m+20}% \right) ^{(-1)^{t_{m}}}\text{.} \end{eqnarray*}$$

and wonder if this form helps to prove the convergence of $P$ in a similar way to the proof of the $\displaystyle\prod_{n=0}^{\infty }\left( 1+b_{n}\right)$ (with $b_{n}\rightarrow 0,b_{n}>0$) is convergent iff $\sum_{n=0}^{\infty }b_{n}$ is convergent.

Question:

A - Which is the Abel's theorem the authors refer to?

B - Suppose we have a product

$$\begin{equation*} P^{\prime }=\displaystyle\prod_{n=0}^{\infty }\left( 1+b_{n}\right) ^{e_{n}}\qquad (\ast ) \end{equation*},$$ where [edited] $b_n>0$ is such that $\displaystyle\sum b_n$ converges [end edit], and $e_{n}$ is a sequence that takes $-1$ and $+1$ values.

1. Is $P^{\prime }$ convergent? I think so for if I take the logarithm of $P$ I get $$\begin{equation*} \sum_{n=0}^{\infty }e_{n}\ln (1+b_{n}) \end{equation*}$$ and this series is absolutely convergent $$\begin{equation*} \sum_{n=0}^{\infty }\left\vert e_{n}\ln (1+b_{n})\right\vert \leq \sum_{n=0}^{\infty }\left\vert \ln (1+b_{n})\right\vert \end{equation*}$$ by the limit comparison test $$\begin{equation*} \underset{n\rightarrow \infty }{\lim }\frac{\ln (1+b_{n})}{b_{n}}=1. \end{equation*}$$
2. Is this argument valid?
3. Similar for $$\begin{equation*} P^{\prime \prime }=\displaystyle\prod_{n=0}^{\infty }\left( 1-a_{n}\right) ^{e_{n}},\qquad (\ast \ast ) \end{equation*},$$ where  $0<a_n<1$ is such that $\displaystyle\sum a_n$ converges [end edit]. The limit comparison test would be $$\begin{equation*} \underset{n\rightarrow \infty }{\lim }\left\vert \frac{\ln (1-a_{n})}{-a_{n}}% \right\vert =1\text{.} \end{equation*}$$
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Yes, provided you replace where $b_n>0$ converges to $0$ by where $b_n>0$ is such that $\sum b_n$ converges and where $a_n>0$ converges to $0$ by where $0<a_n<1$ is such that $\sum a_n$ converges. – Did Apr 25 '11 at 12:00
@Didier Piau, Thanks! I edited the question accordingly. – Américo Tavares Apr 25 '11 at 12:32

To answer question A, this is (what I know under the name of) Abel's method (usually it is for sums, but it works equally well for products of the kind $\prod_n (z_n)^{s_n}$ for $z_n \in \mathbb{C}^{\times}$ and $s_n \in \mathbb{Z}$, or $z_n > 0$ and $s_n \in \mathbb{C}$, or more generally if $z_n \rightarrow 1$ and $s_n$ bounded).
For sums, it allows for example to prove the convergence of $\sum_n a_n b_n$ if $\sum_{k=0}^n b_k$ is bounded and $\sum_n |a_{n+1}-a_n| < \infty$. Abel's method is the discrete counterpart of integration by parts.
In your case, it is easily seen that $S_n = \sum_{k=1}^n s_k$ (where $s_k = (-1)^{t_k}$) is bounded ($s_{2k}+s_{2k+1}=0$), so we write $s_n = S_n - S_{n-1}$, and compute (assuming the $z_n$ are nonzero): $$\begin{eqnarray*} \prod_{n=0}^N z_n^{s_n} & = & \prod_{n=0}^N z_n^{S_n-S_{n-1}} \\ & = & \prod_{n=0}^N z_n^{S_n} \prod_{n=0}^{N-1} z_{n+1}^{-S_n} \\ & = & z_N^{S_N} \prod_{n=0}^{N-1} \left( \frac{z_n}{z_{n+1}} \right)^{S_n} \end{eqnarray*}$$ and in your case $z_n/z_{n+1} = 1 + O(1/n^2)$, so this last product is absolutely convergent (this means that $\sum_n |1-z_n/z_{n+1}| < \infty$, and implies convergence of the product by Cauchy's criterion).
The first 16 terms of the sequence $(s_k=(-1)^{t_k})_{k\ge 0}$ are 1, -1, -1, 1, -1, 1, 1, -1, -1, 1, 1, -1, 1, -1, -1, 1, ... . This is sequence A106400 oeis.org/A106400 – Américo Tavares Apr 25 '11 at 17:38