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.


*

*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
$$\begin{equation*}
\underset{n\rightarrow \infty }{\lim }\frac{\ln (1+b_{n})}{b_{n}}=1.
\end{equation*}$$

*Is this argument valid?

*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 [edit] $0<a_n<1$ is such that $\displaystyle\sum a_n$ converges [end edit]. 
The limit comparison test would be
$$\begin{equation*}
\underset{n\rightarrow \infty }{\lim }\left\vert \frac{\ln (1-a_{n})}{-a_{n}}%
\right\vert =1\text{.}
\end{equation*}$$

 A: 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).
