What is the relation of the Infinite power series with these fraction series?

Question

Take this infinite series: $$S = 1 + \sum_{n=1}^\infty\prod_{i=1}^n\frac{2i+1}{4i} = 1 + \frac{3}{4} + \frac{3\times5}{4\times8} + \frac{3\times5\times7}{4\times8\times12} + ....$$ We want to find the sum of this series. I didn't know how to solve this. But when I went to look at the solution, they compared this series with the infinite power series $$P = (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ......$$ for some real $n$ and $x$. Equating the corresponding terms ($nx = \frac{3}{4}$ and $\frac{n(n-1)}{2}x^2 = \frac{3\times5}{4\times8}$) they found $n=-\frac{3}{2}$ and $x=-\frac{1}{2}$. So that the sum is simply $2^\frac{3}{2}$. And when I checked the infinite power series $P$ plugging these values of $x$ and $n$, it really turn out to be the series $S$. Now, I don't understand why this comparison works. Let's generalise this thing. Say, $S$ is given by $$S = 1 + \sum_{n=1}^\infty\prod_{i=1}^n\frac{ai+b}{di}$$ for some positive integers $a, b$ and $d$ with $b < a$, and it is guaranteed (given) that the series converges. Let $P$ be the same as above. Now, can anyone say if $S$, as defined just now, can always be compared with the series $P$, that is, are there always some real $n$ and $x$ such that $S = P?$

Answer 1

Note that setting $nx=\frac{3}{4}$ and $\frac{n(n-1)}{2!}=\frac{3\cdot 5}{4\cdot 8}$ in order to get values for $n$ and $x$ is just a clever approach, but not a proof that $P=S$.

To make it a proof we have to additionally verify that the general terms evaluated at $n=\frac{3}{2}$ and $x=-\frac{1}{2}$ are equal.

Let's write $S$ and $P$ with general term. We have \begin{align*} S&=1+\frac{3}{4}+\frac{3\cdot 5}{4\cdot 8}+\cdots+ \color{blue}{\frac{3\cdot 5\cdots (2k+1)}{4\cdot 8\cdots (4k)}}+\cdots\tag{1}\\ P&=(1+x)^n\\ &=1+nx+\frac{n(n-1)}{2!}x^2+\cdots+\color{blue}{\frac{n(n-1)\cdots(n-k+1)}{k!}x^k}+\cdots\tag{2} \end{align*} We simplify the general term of $S$ somewhat \begin{align*} \frac{3\cdot 5\cdots (2k+1)}{4\cdot 8\cdots (4k)}&=\frac{1}{4^k}\cdot\frac{3\cdot 5\cdots (2k+1)}{1\cdot 2\cdots k}\\ &\,\,\color{blue}{=\frac{1}{4^kk!}\prod_{j=1}^k(2j+1)} \end{align*} The general term of $P$ evaluated $n=\frac{3}{2}$ and $x=-\frac{1}{2}$: \begin{align*} &\left.\frac{n(n-1)\cdots(n-k+1)}{k!}x^k\right|_{n=-\frac{3}{2},x=-\frac{1}{2}}\\ &\qquad\qquad=\frac{1}{k!}\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\cdots\left(-\frac{3}{2}-k+1\right)\left(-\frac{1}{2}\right)^k\\ &\qquad\qquad=\frac{1}{k!}\frac{(-1)^k}{2^k}\left(3\right)\left(5\right)\cdots\left(3+2k-2\right)\left(-\frac{1}{2}\right)^k\\ &\qquad\qquad\,\,\color{blue}{=\frac{1}{4^kk!}\prod_{j=1}^k(2j+1)} \end{align*} Since both terms are equal we can now conclude that $S=P$.

Note: In order to generalise this approach, we could start to study and compare general terms accordingly.

[Add-on (2020-07-08)]: This add-on is based on OP who has successfully calculated the generalisation in comments to this answer.

The general ($k$-th) term of $S = 1 + \sum_{n=1}^\infty\prod_{j=1}^n\frac{aj+b}{dj}$ is (provided $a\ne 0$): \begin{align*} \prod_{j=1}^k\frac{aj+b}{dj}&=\frac{1}{d^kk!}\prod_{j=1}^k(aj+b)\\ &=\left(\frac{a}{d}\right)^k\frac{1}{k!}\prod_{j=1}^k\left(j+\frac{b}{a}\right)\tag{3} \end{align*} Since the general term of $P=(1+x)^n$ is \begin{align*} \frac{x^k}{k!}\prod_{j=1}^k(n-j+1)&=\frac{(-x)^k}{k!}\prod_{j=1}^{k}\left(j-\left(n+1\right)\right)\tag{4} \end{align*}

we obtain by comparison of (3) with (4): \begin{align*} x=-\frac{a}{d}\qquad\qquad n=-\left(1+\frac{b}{a}\right) \end{align*}

We conclude, providing $|x|=\left|\frac{a}{d}\right|<1$ to assure convergence of the binomial series: \begin{align*} \color{blue}{S }= 1 + \sum_{n=1}^\infty\prod_{j=1}^n\frac{aj+b}{dj}\color{blue}{=\left(1-\frac{a}{d}\right)^{-\left(1+\frac{b}{a}\right)}} \end{align*}

Hints:

• Alternatively to OPs generalisation we can also recall that $P=(1+x)^n$ is some kind of reference series used to derive $S$. We can turn the tables and play with different settings of $x$ and $n$ (respecting $|x|<1$) and check which different general terms and series $S$ we obtain this way.

• A technical aspect. We have to be aware of arithmetic precedence rules and write $\prod_{j=1}^k\color{blue}{(}aj+b\color{blue}{)}$ using brackets, since we have \begin{align*} \prod_{j=1}^kaj+b&=\left(\prod_{j=1}^kaj\right)+b=a^kk!+b\\ \prod_{j=1}^kb+aj&=\left(\prod_{j=1}^kb\right)+aj=b^k+aj\\ \prod_{j=1}^k\left(aj+b\right)&=(a+b)(2a+b)\cdots(ka+b) \end{align*} You might want to see this answer for more information regarding precedence rules.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} S & \equiv 1 + \sum_{n = 1}^{\infty} \prod_{i = 1}^{n}{2i + 1 \over 4i} = 1 + \sum_{n = 1}^{\infty}{2^{n} \over 4^{n}} {\prod_{i = 1}^{n}\pars{i + 1/2} \over n!} = 1 + \sum_{n = 1}^{\infty}{1 \over 2^{n}\, n!}\, \pars{3 \over 2}^{\overline{\large n}} \\[5mm] & = 1 + \sum_{n = 1}^{\infty}{1 \over 2^{n}\, n!}\, {\Gamma\pars{3/2 + n} \over\Gamma\pars{3/2}} = 1 + \sum_{n = 1}^{\infty}{1 \over 2^{n}}\, {\pars{n + 1/2}! \over n!\pars{1/2}!} \\[5mm] & = 1 + \sum_{n = 1}^{\infty}{1 \over 2^{n}}\,{n + 1/2 \choose n} = 1 + \sum_{n = 1}^{\infty}{1 \over 2^{n}} \bracks{{-3/2 \choose n}\pars{-1}^{n}} \\[5mm] & = 1 + \sum_{n = 1}^{\infty} {-3/2 \choose n}\pars{-\,{1 \over 2}}^{n} = 1 + \braces{\bracks{1 + \pars{-\,{1 \over 2}}}^{-3/2} - 1} \\[5mm] & = \bbox[15px,#ffd,border:1px solid navy]{\large 2\root{2}}\ \approx\ 2.8284 \end{align}