 Duncan Reed

2022-01-21

A series is given as follows
$\frac{1}{6}+\frac{5}{6\cdot 12}+\frac{5\cdot 8}{6\cdot 12\cdot 18}+\frac{5\cdot 8\cdot 11}{6\cdot 12\cdot 18\cdot 24}+\dots$ rakije2v

Expert

$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2}\prod _{i=1}^{n}\frac{3i-1}{6i}=\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{2}^{n}n!}\prod _{i=1}^{n}\frac{3i-1}{3}$
$=\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\left(-2\right)}^{n}n!}\prod _{i=1}^{n}\frac{1-3i}{3}$
$=\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\left(-2\right)}^{n}n!}\prod _{i=1}^{n}\left(\frac{1}{3}-i\right)$
$={\left[\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{{x}^{n}}{n!}\prod _{i=1}^{n}\left(\frac{1}{3}-i\right)\right]}_{x=-\frac{1}{2}}$
Now note that for , where $f\left(x\right)={\left(x+1\right)}^{-\frac{2}{3}}$. We can manipulate the expression a little further to see it as a Taylor series for f.
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2}\prod _{i=1}^{n}\frac{3i-1}{6i}={\left[\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}\frac{{x}^{n}}{n!}\prod _{i=1}^{n}\left(\frac{1}{3}-i\right)-\frac{1}{2}\right]}_{x=-\frac{1}{2}}$
$={\left[\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}\frac{{x}^{n}}{n!}{f}^{\left(n\right)}\left(0\right)\right]}_{x=-\frac{1}{2}}-\frac{1}{2}$
$=\frac{1}{2}f\left(-\frac{1}{2}\right)$ RizerMix

Expert

Consider the binomial expansion of $\left(1+x{\right)}^{n}=1+nx+\frac{n\left(n-1\right)}{2}{x}^{2}+...$ Multiplying the series term by term by 2 On comparing, $nx=\frac{1}{3}$ and $\frac{n\left(n-1\right)}{2}{x}^{2}=\frac{5}{3.12}$ Solving for n,x we get So the sum becomes $\left[\left(\frac{1}{2}{\right)}^{\frac{-2}{3}}-1\right]={2}^{\frac{2}{3}}-1={4}^{\frac{1}{3}}-1$ since 2 has been multiplied before divide the result by 2 nick1337

Expert

For any $n\in {\mathbb{N}}^{+}$ we have
$\prod _{k=1}^{n}\left(3k+2\right)=\frac{{3}^{n}\mathrm{\Gamma }\left(n+\frac{5}{3}\right)}{\mathrm{\Gamma }\left(\frac{5}{3}\right)}$
and the value of the RHS at n=0 equals 1. It follows that the whole series can be represented as
$S=\sum _{n\ge 0}\frac{1}{{6}^{n+1}\left(n+1\right)!}\cdot \frac{{3}^{n}\mathrm{\Gamma }\left(n+\frac{5}{3}\right)}{\mathrm{\Gamma }\left(\frac{5}{3}\right)}$
$=\sum _{n\ge 0}\frac{1}{6\cdot {2}^{n}}\cdot \frac{\mathrm{\Gamma }\left(n+\frac{5}{3}\right)}{\mathrm{\Gamma }\left(\frac{5}{3}\right)\mathrm{\Gamma }\left(n+2\right)}$
or, by using Euler's beta function and the reflection formula for the $\mathrm{\Gamma }$ function, as
$\sum _{n\ge 0}\frac{1}{6\mathrm{\Gamma }\left(\frac{1}{3}\right)\mathrm{\Gamma }\left(\frac{5}{3}\right)}\cdot \frac{B\left(n+\frac{5}{3},\frac{1}{3}\right)}{{2}^{n}}=\frac{\sqrt{3}}{8\pi }\sum _{n\ge 0}{\int }_{0}^{1}\frac{{x}^{n+2/3}\left(1-x{\right)}^{-2/3}}{{2}^{n}}dx$
from which:
$S=\frac{\sqrt{3}}{4\pi }{\int }_{0}^{1}\frac{{x}^{2/3}}{\left(1-x{\right)}^{2/3}\left(2-x\right)}dx=\frac{\sqrt{3}}{4\pi }{\int }_{0}^{+\mathrm{\infty }}\frac{{z}^{2/3}}{\left(1+z\right)\left(2+z\right)}$
and:
$S=\frac{\sqrt{3}}{4\pi }\left(-{\int }_{0}^{+\mathrm{\infty }}\frac{{z}^{2/3}}{z\left(1+z\right)}dz+2{\int }_{0}^{+\mathrm{\infty }}\frac{{z}^{2/3}}{z\left(2+z\right)}dz\right)$
Euler's beta function then leads to:
$S=\frac{1}{\sqrt{2}}-\frac{1}{2}\approx 0.2937$

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