3

How to find sum of above series

$$\frac{1}{6} +\frac{5}{6\cdot12} +\frac{5\cdot8}{6\cdot12\cdot18} +\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+...$$ How to find sum of series I can find its convergence but not sum of series.

Can anyone explain?

OR.
  • 5,941
Ni TiSh
  • 233

4 Answers4

6

Like Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $

as $$\dfrac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}=\dfrac32\cdot\dfrac{-2/3\cdot-5/3\cdot-8/3\cdot-11/3}{4!}\left(-\dfrac36\right)^4$$

$$\dfrac16=\dfrac32\cdot\dfrac{-2/3}{1!}\left(-\dfrac36\right)^1$$

So, the sum $$=-1+\dfrac32\cdot\left(1-\dfrac36\right)^{-2/3}$$

lab bhattacharjee
  • 274,582
3

Using binomial expansion \begin{eqnarray*} (1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots \end{eqnarray*} with $x=-\frac{1}{2}$ and $n=-\frac{2}{3}$. \begin{eqnarray*} (1-\frac{1}{2})^{-\frac{2}{3}}=1+\left(-\frac{2}{3}\right)\left(-\frac{1}{2}\right)+\frac{\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{2}\left(-\frac{1}{2}\right)^2+\frac{\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)\left(-\frac{8}{3}\right)}{3!}\left(-\frac{1}{2}\right)^3+\cdots \end{eqnarray*} Rearrange a bit & we have \begin{eqnarray*} \color{red}{\frac{2^{\frac{2}{3}}-1}{2}}=\frac{1}{6} +\frac{5}{6\cdot12} +\frac{5\cdot8}{6\cdot12\cdot18} +\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\cdots \end{eqnarray*}

Donald Splutterwit
  • 36,613
  • 2
  • 26
  • 73
3

We want to compute: $$ S = \sum_{n\geq 1}\frac{1}{6^n n!}\cdot\frac{1}{2}\prod_{k=1}^{n}(3k-1) = \sum_{n\geq 1}\frac{\Gamma\left(n+\frac{2}{3}\right)}{2^{n+1}\,\Gamma\left(\frac{2}{3}\right)\,\Gamma(n+1)}=\sum_{n\geq 1}\frac{B\left(n+\frac{2}{3},\frac{1}{3}\right)}{2^{n+1}\cdot\frac{2\pi}{\sqrt{3}}}$$ that by Euler's Beta function equals $$ \frac{\sqrt{3}}{4\pi}\sum_{n\geq 1}\int_{0}^{1}\frac{1}{2^n} x^{n-1/3}(1-x)^{-2/3}\,dx =\frac{\sqrt{3}}{4\pi}\int_{0}^{1}\frac{x^{2/3}}{(1-x)^{2/3}(2-x)}\,dx$$ or $$ \frac{\sqrt{3}}{4\pi}\int_{0}^{1}\frac{(1-x)^{2/3}}{x^{2/3}(1+x)}\,dx =\frac{3\sqrt{3}}{4\pi}\int_{0}^{1}\frac{(1-x^3)^{2/3}}{1+x^3}\,dx=\color{red}{\frac{2^{2/3}-1}{2}}.$$

Jack D'Aurizio
  • 353,855
  • 1
    Sir how u think about using eular beta function.how to approach that question what logic u have used – Ni TiSh Jun 18 '17 at 14:23
  • 1
    @NiTiSh: The Euler Beta function allows an exact evaluation of many hypergeometric series, like $$ \sum_{n\geq 1}\frac{1}{n^2 2^n \binom{3n}{n}}=\frac{\pi^2}{24}-\frac{\log^2(2)}{2}$$ (I am going to publish an article about the application of Euler's Beta function and the dilogarithm for computing some values of $\phantom{}_{4} F_3$), and as a rule of thumb, it always deserves a try. – Jack D'Aurizio Jun 18 '17 at 14:42
  • That said, it is not the unique technique for tackling the given problem. The generalized binomial theorem and Lagrange's inversion formula are also pretty effective here. – Jack D'Aurizio Jun 18 '17 at 14:43
  • Have a look at Achille Hui's answer here: https://math.stackexchange.com/a/1600431/44121 – Jack D'Aurizio Jun 18 '17 at 14:45
  • For the downvoter(s): please explain your downvote. – Jack D'Aurizio Jun 20 '17 at 15:54
3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &{1 \over 6} + {5 \over 6 \cdot 12} + {5 \cdot 8 \over 6 \cdot 12 \cdot 18} + {5 \cdot 8 \cdot 11 \over 6 \cdot 12 \cdot 18 \cdot 24} + \cdots \equiv {1 \over 6} + \sum_{n = 0}^{\infty}{\prod_{k = 0}^{n}\pars{3k + 5} \over \prod_{k = 0}^{n + 1}\pars{6k + 6}} \\[5mm] = &\ {1 \over 6} + \sum_{n = 0}^{\infty} {3^{n + 1}\prod_{k = 0}^{n}\pars{k + 5/3} \over 6^{n + 2}\prod_{k = 0}^{n + 1}\pars{k + 1}} = {1 \over 6} + {1 \over 12}\sum_{n = 0}^{\infty} {\pars{5/3}^{\overline{n + 1}} \over2^{n}\pars{n + 2}!} \\[5mm] = &\ {1 \over 6} + {1 \over 12}\sum_{n = 0}^{\infty} {\Gamma\pars{8/3 + n}/\Gamma\pars{5/3} \over2^{n}\pars{n + 2}!} = {1 \over 6} + {1 \over 12\,\Gamma\pars{5/3}}\sum_{n = 0}^{\infty} {\pars{n + 5/3}! \over 2^{n}\pars{n + 2}!} \\[5mm] = &\ {1 \over 6} + {\pars{-1/3}! \over 12\,\Gamma\pars{5/3}}\sum_{n = 0}^{\infty} {\pars{n + 5/3}! \over \pars{n + 2}!\,\pars{-1/3}!}\,\pars{1 \over 2}^{n} \\[5mm] = &\ {1 \over 6} + {\Gamma\pars{2/3} \over 12\bracks{\pars{2/3}\Gamma\pars{2/3}}}\sum_{n = 0}^{\infty} {n + 5/3 \choose n + 2}\pars{1 \over 2}^{n} = {1 \over 6} + {1 \over 8} \sum_{n = 0}^{\infty}{-2/3 \choose n + 2}\pars{-1}^{n + 2}\pars{1 \over 2}^{n} \\[5mm] = &\ {1 \over 6} + {1 \over 8} \sum_{n = 2}^{\infty}{-2/3 \choose n}\pars{-\,{1 \over 2}}^{n - 2} \\[5mm] = &\ {1 \over 6} + {1 \over 2}\braces{\bracks{1 + \pars{-\,{1 \over 2}}}^{-2/3} - {-2/3 \choose 0} - {-2/3 \choose 1}\pars{-\,{1 \over 2}}} = {1 \over 6} + {1 \over 2}\pars{2^{2/3} - 1 - {1 \over 3}} \\[5mm] = &\ \bbx{2^{2/3} - 1 \over 2} \end{align}

Felix Marin
  • 89,464