Convergence of the sum of products $\sum_{k=0}^\infty \prod_{j=1}^k \left(1-\frac{3}{2j}\right)$

Question

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?

Answer 1

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 $\sum_{k=0}^\infty k^{-3/2}$.

Answer 2

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.

Answer 3

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 $$\log(S_p)=-\frac 12 \log(\pi p)-\frac{1}{8 p}+\frac{1}{192 p^3}+O\left(\frac{1}{p^5}\right)$$ $$S_p=e^{\log(S_p)}\sim \frac 1 {\sqrt{\pi p}} \exp\left(-\frac{1}{8 p} \right)$$

Answer 4

I've found a more elementary way to prove that this series converges absolutely, using the comparison test.
Let $a_k=\prod_{j=1}^k \frac{2j-3}{2j} = \frac{-1\cdot1\cdot3\cdot5\cdot...\cdot(2k-3)}{2\cdot4\cdot6\cdot8\cdot...\cdot2k}$ and $c_k=\prod_{j=1}^k \frac{2j-1}{2j} = \frac{1\cdot3\cdot5\cdot...\cdot(2k-3)\cdot(2k-1)}{2\cdot4\cdot6\cdot8\cdot...\cdot2k} $ for $k\in\Bbb{N} $. $$ |a_k|=\left\vert\frac{-1\cdot1\cdot3\cdot5\cdot...\cdot(2k-3)}{2\cdot4\cdot6\cdot8\cdot...\cdot2k}\right\vert = \frac{1\cdot1\cdot3\cdot5\cdot...\cdot(2k-3)}{2\cdot4\cdot6\cdot8\cdot...\cdot2k} = \frac{\prod_{j=2}^k 2j-3}{\prod_{j=1}^k 2j} = \frac{\prod_{j=1}^{k-1} 2j-1}{\prod_{j=1}^k 2j} $$ so $$ |a_k|(2k-1)=\frac{(2k-1)\prod_{j=1}^{k-1} 2j-1}{\prod_{j=1}^k 2j}=\frac{\prod_{j=1}^{k} 2j-1}{\prod_{j=1}^k 2j}=c_k $$ We can use induction to prove that $c_k \leq \frac{1}{\sqrt{3k+1}} $ for any $k\in\Bbb{N}$: when $k=1, c_1=\frac{1}{\sqrt{3\cdot1+1}}$; when $k=2, c_2=\frac 3 8 =.375 $ while $\frac{1}{\sqrt{3\cdot2+1}} \approx.377964$; and for any $k\in\Bbb{N}$, if $c_k\lt \frac{1}{\sqrt{3k+1}} $, then $c_{k+1}=c_k\cdot\frac{2k+1}{2k+2}\lt\frac{2k+1}{(2k+2)\cdot\sqrt{3k+1}} $ and $$\left(\frac{2k+1}{(2k+2)\cdot\sqrt{3k+1}}\right)^2=\frac{(2k+1)^2}{(2k+2)^2\cdot(3k+1)}=\frac{(2k+1)^2}{12k^3+28k^2+20k+4}=\frac{(2k+1)^2}{12k^3+28k^2+19k+4+k}=\frac{(2k+1)^2}{(2k+1)^2(3k+4)+k}\lt\frac{(2k+1)^2}{(2k+1)^2(3k+4)}=\frac{1}{3k+4}=\frac{1}{3(k+1)+1} $$ so $\frac{2k+1}{(2k+2)\cdot\sqrt{3k+1}}\lt\frac{1}{\sqrt{3(k+1)+1}} $. (The details of this proof are here.)

Therefore $|a_k|(2k-1) \leq \frac{1}{\sqrt{3k+1}} \lt \frac{1}{\sqrt{3k}} $, so $|a_k|\lt\frac{1}{(2k-1)\sqrt{3k}} $.
$2k-1\geq k$ for any $k\in\Bbb{N}$, so $\frac{1}{(2k-1)\sqrt{3k}}\leq\frac{1}{k\sqrt{3k}}=\frac{1}{\sqrt{3}\cdot k^{3/2}} $. Put this into the series, and
$ \sum_{k=1}^\infty |a_k| \lt \frac{1}{\sqrt3}\sum_{k=1}^\infty \frac{1}{k^{3/2}} $, which, being a P-series with an exponent greater than 1, converges, so
$ \sum_{k=0}^\infty \prod_{j=1}^k \frac{2j-3}{2j} = 1+\sum_{k=1}^\infty a_k $ converges absolutely.
$\sum_{k=0}^\infty \binom{1/2}k (x^2-1)^2$ should also converge absolutely when $x=\pm\sqrt2$, because that produces a series with the same terms except that the terms' sign alternates because $(x^2-1)^k=1^k$ rather than $(-1)^k$.

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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}