# Sum of series $\frac{ 4}{10}+\frac{4\cdot 7}{10\cdot 20}+\frac{4\cdot 7 \cdot 10}{10\cdot 20 \cdot 30}+\cdots \cdots$

Sum of the series

$$\frac {4}{10}+\frac {4\cdot7}{10\cdot20}+ \frac {4\cdot7\cdot10}{10\cdot20\cdot30}+\cdots$$

Sum of series $\frac {4}{10}+\frac {4\cdot7}{10\cdot20}+ \frac {4\cdot7\cdot10}{10\cdot20\cdot30}+\cdots$

But i want to solve it using Beta ganmma function.

My Process follows

Let $$S = \frac{1}{10}\bigg[\frac{ 4}{1}+\frac{4\cdot 7}{1\cdot 2}+\frac{4\cdot 7 \cdot 10}{1\cdot 2 \cdot 3}+\cdots \cdots \bigg]$$

Let $\displaystyle a_{n} = \prod^{n}_{k=1}(3k+1)=3^n \prod^{n}_{k=1}\bigg(k+\frac{1}{3}\bigg)$

$\displaystyle =3^n\Gamma\bigg(n+\frac{4}{3}\bigg)\cdot \frac{1}{\Gamma \bigg(\frac{4}{3}\bigg)}$

And Let $\displaystyle b_{n} = \prod^{n}_{k=1}k = \Gamma (n+1)$

So $\displaystyle S =\frac{1}{10} \sum^{\infty}_{n=1}3^n \frac{\Gamma \bigg(n+\frac{4}{3}\bigg)\cdot \Gamma \bigg(\frac{2}{3}\bigg)}{\Gamma \bigg(\frac{2}{3}\bigg)\Gamma \bigg(\frac{4}{3}\bigg)\cdot \Gamma(n+1)}$

using $\displaystyle \Gamma(x)\cdot \Gamma(1-x) = \frac{\pi}{\sin (\pi x)}$ and

$\displaystyle \frac{\Gamma (a) \Gamma(b)}{\Gamma(a+b)} =B(a,b)= \int^{1}_{0}x^{a-1}(1-x)^{b-1}dx$

So $\displaystyle S = \frac{1}{10}\sum^{\infty}_{n=1}3^n\int^{1}_{0}x^{n+\frac{1}{3}}(1-x)^{-\frac{1}{3}}dx$

$\displaystyle =\frac{1}{10}\int^{1}_{0}\frac{3x}{1-3x}\cdot \bigg(\frac{x}{1-x}\bigg)^{\frac{1}{3}}dx$

Now put $\displaystyle \frac{x}{1-x} = t\Rightarrow x = \frac{t}{1+t} = 1-\frac{1}{1+t}$

So $\displaystyle S = \frac{3}{10}\int^{\infty}_{0}\frac{t^{\frac{4}{3}}}{(2t-1)(1+t)^2}dt$

Could some Help me to solve it, Thanks

although it has solved Here

• It is trivially divergent since the main term of the series is greater than $3^n$. – Jack D'Aurizio Sep 19 '18 at 7:27
• @JackD'Aurizio I think the OP forgot zeros. He said he wanted to calculate $$\frac {4}{10}+\frac {4\cdot7}{10\cdot20}+ \frac {4\cdot7\cdot10}{10\cdot20\cdot30}+\cdots$$ (see the link he gave) using the beta and gamma functions. I think it is a bit premature to close the thread now. – Batominovski Sep 19 '18 at 7:29
• Sorry friends i have edited my post and want solution for question which Batominovski statet using beta gamma function – DXT Sep 19 '18 at 7:52
• @Batominovski: nevertheless, it is more or less the same question asked over and over, which is how to use the Beta function for computing $\sum_{n\geq 1}\prod_{k=1}^n\frac{Ak+B}{Ck+D}$. – Jack D'Aurizio Sep 19 '18 at 11:23

For a complex number $z$ such that $|z|<1$, let $$f(z):=\sum_{n=0}^\infty\,\prod_{k=1}^n\,\left(\frac{k+\frac{1}{3}}{k}\right)\,z^n\,.$$ Then, the required sum $S:=\displaystyle\sum_{n=1}^\infty\,\prod_{k=1}^n\,\left(\frac{3k+1}{10k}\right)$ satisfies $$S=f\left(\frac{3}{10}\right)-1\,.$$ We claim that $f(z)=(1-z)^{-\frac43}$ for all complex numbers $z$ such that $|z|<1$. As a consequence, $$S=\left(\frac{7}{10}\right)^{-\frac43}-1=\left(\frac{10}{7}\right)^{\frac43}-1\approx0.608926\,.$$

Note that, for $n=0,1,2,3,\ldots$, \begin{align}t_n&:=\prod_{k=1}^n\,\left(\frac{k+\frac{1}{3}}{k}\right)\\&=\frac{\Gamma\left(n+\frac43\right)}{\Gamma\left(\frac43\right)\,\Gamma(n+1)}\\&=\frac{(n+1)\,\Gamma\left(n+\frac43\right)}{\Gamma\left(\frac43\right)\,\Gamma(n+2)}\,,\end{align} where $\Gamma$ is the usual gamma function. That is, $$t_n=\frac{n+1}{\Gamma\left(\frac43\right)\,\Gamma\left(\frac23\right)}\,\text{B}\left(n+\frac43,\frac23\right)\,,$$ where $\text{B}$ is the usual beta function. From the identity $$\Gamma(x)\,\Gamma(1-x)=\frac{\pi}{\sin(\pi\,x)}\,,$$ we conclude that $$\Gamma\left(\frac43\right)\,\Gamma\left(-\frac13\right)=\frac{\pi}{\sin\left(\frac{4\pi}{3}\right)}=-\frac{2\pi}{\sqrt{3}}\,.$$ Ergo, $$\Gamma\left(\frac43\right)\,\Gamma\left(\frac23\right)=\left(-\frac13\right)\,\left(-\frac{2\pi}{\sqrt{3}}\right)=\frac{2\pi}{3\sqrt{3}}\,.$$ In other words, $$t_n=\frac{3\sqrt{3}}{2\pi}\,(n+1)\,\int_0^1\,u^{n+\frac13}\,(1-u)^{-\frac13}\,\text{d}u\,.$$

We now obtain $$f(z)=\frac{3\sqrt{3}}{2\pi}\,\int_0^1\,\left(\sum_{n=0}^\infty\,(n+1)\,u^{n+\frac13}\,z^n\right)\,(1-u)^{-\frac13}\,\text{d}u\,.$$ This gives $$f(z)=\frac{3\sqrt{3}}{2\pi}\,\int_0^1\,\frac{u^{\frac13}\,(1-u)^{-\frac13}}{(1-zu)^2}\,\text{d}u\,.$$ Let $v:=\dfrac{u}{1-u}$. Then, $$f(z)=\frac{3\sqrt{3}}{2\pi}\,\int_0^\infty\,\frac{v^{\frac13}}{\big(1+(1-z)v\big)^2}\,\text{d}v\,.$$

The integral $$J:=\int_0^\infty\,\frac{v^{\frac13}}{\big(1+(1-z)v\big)^2}\,\text{d}v$$ can be evaluated using the same keyhole contour as in this thread. It is not difficult to see that \begin{align}J&=\left(\frac{2\pi\text{i}}{1-\exp\left(\frac{2\pi\text{i}}{3}\right)}\right)\,\text{Res}_{v=-\frac{1}{1-z}}\left(\frac{v^{\frac13}}{\big(1+(1-z)v\big)^2}\right) \\&=-\frac{\pi\,\exp\left(-\frac{\pi\text{i}}{3}\right)}{\sin\left(\frac{\pi}{3}\right)}\,\Biggl(-\frac{\exp\left(\frac{\pi\text{i}}{3}\right)}{3}\,(1-z)^{-\frac43}\Biggr) \\&=\frac{2\pi}{3\sqrt{3}}\,(1-z)^{-\frac{4}{3}}\,.\end{align} That is, $$f(z)=(1-z)^{-\frac43}\text{ for all }z\in\mathbb{C}\text{ such that }|z|<1\,.$$

Well, the OP requested a long and combersome solution. The result can be easily proven via the binomial theorem: $$(1-z)^{-\frac43}=\sum_{n=0}^\infty\,\binom{-\frac43}{n}\,(-1)^n\,z^n=\sum_{n=0}^\infty\,\prod_{k=1}^n\,\left(\frac{k+\frac13}{k}\right)\,z^n$$ for every complex number $z$ with $|z|<1$.

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

$\ds{{4 \over 10} + {4 \cdot 7 \over 10 \cdot 20} + {4 \cdot 7 \cdot 10 \over 10 \cdot 20 \cdot 30} + \cdots = \sum_{n = 1}^{\infty}{\prod_{k = 1}^{n}\pars{3k + 1} \over \prod_{k = 1}^{n}10k}:\ {\LARGE ?}}$.

\begin{align} &\bbox[10px,#ffd]{\ds{\sum_{n = 1}^{\infty} {\prod_{k = 1}^{n}\pars{3k + 1} \over \prod_{k = 1}^{n}10k}}} = \sum_{n = 1}^{\infty} {3^{n}\prod_{k = 1}^{n}\pars{k + 1/3} \over 10^{n}n!} = \sum_{n = 1}^{\infty} {\pars{3/10}^{n} \over n!}\,\pars{4 \over 3}^{\large\overline{n}} \\[5mm] = &\ \sum_{n = 1}^{\infty} {\pars{3/10}^{n} \over n!}\,{\Gamma\pars{4/3 + n} \over \Gamma\pars{4/3}} = \sum_{n = 1}^{\infty} \pars{3 \over 10}^{n}\,{\pars{n + 1/3}! \over n!\pars{1/3}!} \\[5mm] = &\ \sum_{n = 1}^{\infty}\pars{3 \over 10}^{n}\,{n + 1/3 \choose n} = \sum_{n = 1}^{\infty}\pars{3 \over 10}^{n} \bracks{{-4/3 \choose n}\pars{-1}^{n}} \\[5mm] = &\ \sum_{n = 1}^{\infty} {-4/3 \choose n}\pars{-\,{3 \over 10}}^{n} = \bracks{1 + \pars{-\,{3 \over 10}}}^{-4/3} - 1 \\[5mm] = &\ \bbx{\pars{10 \over 7}^{4/3} - 1} \approx 0.6089 \end{align}