Convergence of the sum of products $\sum_{k=0}^\infty \prod_{j=1}^k \left(1-\frac{3}{2j}\right)$ How can I prove that $\sum_{k=0}^\infty \prod_{j=1}^k \left(1-\frac{3}{2j}\right)$ converges?
I'm trying to prove that a polynomial approximation of the absolute value function converges. I know from the generalized binomial theorem that
$|x| = ((x^2-1)+1)^{1/2} = \sum_{k=0}^\infty {1/2\choose k} (x^2-1)^k = \sum_{k=0}^\infty \left(\prod_{j=1}^k \frac{3-2j}{2j}\right)(x^2-1)^k$ which converges when $|x^2-1|<1$, i.e. $0<x<\sqrt{2}$. However, when $x=0$, the series is
\begin{multline*}
\sum_{k=0}^\infty \left(\prod_{j=1}^k \frac{3-2j}{2j}\right)(-1)^k = \sum_{k=0}^\infty (-1)^{k}\prod_{j=1}^k \left(\frac{3}{2j}-1\right) \\ = \sum_{k=0}^\infty (-1)^{k}\prod_{j=1}^k \left(-\left(1-\frac{3}{2j}\right)\right) = \sum_{k=0}^\infty \prod_{j=1}^k \left(1-\frac{3}{2j}\right).
\end{multline*}
I know that this should converge (my textbook uses the fact that $\sum_{k=0}^\infty {1/2\choose k} (x^2-1)^k$ converges when $|x|<1$ to prove a different theorem), but how can I prove that it converges?
 A: Let
$$a_k=\prod_{j=1}^k\left(1-\frac{3}{2j}\right).$$
Then
$$\ln a_k=\sum_{j=1}^k\left(1-\frac{3}{2j}\right)
=\sum_{j=1}^k\left(-\frac{3}{2j}+O(j^{-2})\right)=-\frac32\ln k+O(1).$$
So
$$a_k\le Ck^{-3/2}$$
for some $C$, and $\sum_{k=0}^\infty a_k$
converges by comparison to $\sum_{k=0}^\infty k^{-3/2}$.
A: For all $t$ such that $|t| < 1,$
$$
1 - \sqrt{1 - t} = \sum_{k=1}^\infty(-1)^{k-1}\binom{\frac12}kt^k
= \sum_{k=1}^\infty b_kt^k,
$$
where
$$
b_k = \left\lvert\binom{\frac12}k\right\rvert =
\frac12\cdot\prod_{j=2}^k\frac{2j - 3}{2j} \quad (k \geqslant 1).
$$
Define
$$
c_k = (2k - 1)b_k = \prod_{j=1}^k\frac{2j - 1}{2j} \quad (k \geqslant 1).
$$
At this point, noting that $(2k - 1)b_k < 1,$ we could apply
Littlewood's Tauberian theorem. Alternatively, noting that
$b_k > 0,$ we could apply the Tauberian theorem given as Exercise
9.37 in Apostol, Mathematical Analysis (2nd ed. 1974). But it is
enough to apply Tauber's first theorem without elaboration, because:
\begin{align*}
c_k & = \prod_{j=1}^k\left(1 - \frac1{2j}\right)
< \left(\prod_{j=1}^k\left(1 + \frac1{2j}\right)\right)^{-1} \!\!
< \left(1 + \sum_{j=1}^k\frac1{2j}\right)^{-1} \!\!
\to 0 \text{ as } k \to \infty,
\end{align*}
therefore
$$
b_k = o\left(\frac1k\right).
$$
Here is Tauber's first theorem, as given by Apostol (p.246f.):

Theorem 9.33 (Tauber).
Let $f(x) = \sum_{n=0}^\infty a_nx^n$ for $ -1 < x < -1,$ and assume
that $\lim_{n \to \infty}na_n = 0.$ If $f(x) \to S$ as $x \to 1-,$
then $\sum_{n=0}^\infty a_n$ converges and has sum $S.$

In the present instance, we have
$$
1 - \sum_{k=1}^\infty b_kt^k = \sqrt{1 - t} \to 0 \text{ as }
t \to 1-,
$$
and $\lim_{k \to \infty} kb_k = 0,$ therefore
$$
1 - \sum_{k=1}^\infty\left\lvert\binom{\frac12}k\right\rvert =
1 - \sum_{k=1}^\infty b_k = 0,
$$
as required.
A: As already said in comments $$P_k=\prod_{j=1}^k \left(1-\frac{3}{2j}\right)=-\frac{1}{2 \sqrt{\pi }}\frac{\Gamma \left(k-\frac{1}{2}\right)}{ \Gamma (k+1)}$$ Considering the partial sums
$$S_p=\sum_{k=0}^p P_k$$, this generates the sequence
$$\left\{1,\frac{1}{2},\frac{3}{8},\frac{5}{16},\frac{35}{128},\frac{63}{256},\frac{2
   31}{1024},\frac{429}{2048},\frac{6435}{32768},\frac{12155}{65536},\frac{46189}{2
   62144},\cdots\right\}$$
The numerators correspond to sequence $A001790$ in $OEIS$ (they are the numerators in the expansion of $\frac{1}{\sqrt{1-x}}$).
The denominators correspond to sequence $A046161$ in $OEIS$ (they are the  denominators of $4^{-n} \binom{2 n}{n}$).
As a result, we have
$$S_p= \frac{\Gamma \left(p+\frac{1}{2}\right)}{\sqrt{\pi } \, \Gamma (p+1)}$$ Using Stirling approximation and continuing with Taylor expansions
$$\log(S_p)=-\frac 12 \log(\pi p)-\frac{1}{8
   p}+\frac{1}{192 p^3}+O\left(\frac{1}{p^5}\right)$$
$$S_p=e^{\log(S_p)}\sim \frac 1 {\sqrt{\pi p}} \exp\left(-\frac{1}{8
   p} \right)$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
I'll assume that the sum over $\ds{k}$ starts at $\ds{\Large\color{red}{1}}$. Namely
\begin{align}
&\bbox[5px,#ffd]{\sum_{k =\color{red}{\Large 1}}^{\infty}\prod_{j = 1}^{k}\pars{1 - {3 \over 2j}}} =
\sum_{k = 1}^{\infty}\prod_{j = 1}^{k}{j - 3/2 \over j} =
\sum_{k = 1}^{\infty}{\pars{-1/2}^{\overline{k}} \over k!}
\\[5mm] = &\
\sum_{k = 1}^{\infty}{\Gamma\pars{-1/2 + k}/\Gamma\pars{-1/2} \over k!} =
\sum_{k = 1}^{\infty}{\pars{k - 3/2}! \over k!\pars{-3/2}!} =
\sum_{k = 1}^{\infty}{k - 3/2 \choose k}
\\[5mm] = &\
\sum_{k = 1}^{\infty}{1/2 \choose k}\pars{-1}^{k} =
\bracks{1 + \pars{-1}}^{1/2} - {1/2 \choose 0}\pars{-1}^{0} = \bbx{\large -1} \\ &
\end{align}
