Sum $\frac{1}{6} + \frac{5}{6\cdot 12} + \frac{5\cdot 8}{6\cdot 12\cdot 18} + \frac{5\cdot 8\cdot 11}{6\cdot 12\cdot 18\cdot 24}+\ldots$ [closed]

A series is given as follows

$$\frac{1}{6} + \frac{5}{6\cdot 12} + \frac{5\cdot 8}{6\cdot 12\cdot 18} + \frac{5\cdot 8\cdot 11}{6\cdot 12\cdot 18\cdot 24}+\ldots$$

Can you give me hints to get started finding its value? Thanks.

• 6.12.18, are you sure? – Jorge Fernández Hidalgo Jan 5 '15 at 5:20
• Can you please edit your thread to use LaTeX? It will make things much clearer. – ml0105 Jan 5 '15 at 5:20
• It would be nice if you would properly format your question. – Gahawar Jan 5 '15 at 5:20
• I think the intent is $\sum_{n=1}^\infty \frac{\prod_{i=0}^{n-2} 5+3i}{\prod_{i=1}^n 6i}$, where my convention is that the empty product $\prod_{i=0}^{-1} a_i$ is $1$. – Ian Jan 5 '15 at 5:25
• You can recognize this series as a Taylor series for $\frac12(x+1)^{-2/3}-\frac12$ evaluated at $x=-{\frac12}$. The general way to handle this kind of series is to (1) recognize that the denominator has $n!$ that can be factored out, which is part of the taylor formula, and (2) recognize the numerator's factors increase by $3$. So with appropriate manipulation, you can find a series for a function of the form $(1+x)^{k/3}$ and an $x$-value that almost gives you your series. (There is also a scaling factor and an accounting of the constant term to work out.) – alex.jordan Jan 5 '15 at 6:28

Consider the binomial expansion of $(1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+...$ Multiplying the series term by term by $2$

On comparing, $nx=\frac{1}{3}$ and $\frac{n(n-1)}{2}x^2=\frac{5}{3.12}$

Solving for $n,x$ we get $n=\frac{-2}{3},x=\frac{-1}{2}$

So the sum becomes $\{(\frac{1}{2})^{\frac{-2}{3}}-1\}=2^\frac{2}{3}-1=4^\frac{1}{3}-1$

since $2$ has been multiplied before divide the result by $2$

• thanks for answer , but answer is what as stated by oleg567 in comment – Sophie Clad Jan 5 '15 at 5:51
• @SophieClad check if this suits you – Learnmore Jan 5 '15 at 6:26

\begin{align} \sum_{n=1}^{\infty}\frac12\prod_{i=1}^n\frac{3i-1}{6i} &=\frac12\sum_{n=1}^{\infty}\frac1{2^nn!}\prod_{i=1}^n\frac{3i-1}{3}\\ &=\frac12\sum_{n=1}^{\infty}\frac1{(-2)^nn!}\prod_{i=1}^n\frac{1-3i}{3}\\ &=\frac12\sum_{n=1}^{\infty}\frac1{(-2)^nn!}\prod_{i=1}^n\left(\frac13-i\right)\\ &=\left[\frac12\sum_{n=1}^{\infty}\frac{x^n}{n!}\prod_{i=1}^n\left(\frac13-i\right)\right]_{x=-1/2}\\ \end{align}

Now note that for $n\geq1$, $\prod_{i=1}^n\left(\frac13-i\right)=f^{(n)}(0)$, where $f(x)=(x+1)^{-2/3}$. We can manipulate the expression a little further to see it as a Taylor series for $f$.

\begin{align} \sum_{n=1}^{\infty}\frac12\prod_{i=1}^n\frac{3i-1}{6i} &=\left[\frac12\sum_{n=0}^{\infty}\frac{x^n}{n!}\prod_{i=1}^n\left(\frac13-i\right)-\frac12\right]_{x=-1/2}\\ &=\left[\frac12\sum_{n=0}^{\infty}\frac{x^n}{n!}f^{(n)}(0)\right]_{x=-1/2}-\frac12\\ &=\frac12f(-1/2)-\frac12\\ &=\frac12(1/2)^{-2/3}-\frac12\\ &=\sqrt[3]{1/2}-\frac12\\ \end{align}

For any $n\in\mathbb{N}^+$ we have $$\prod_{k=1}^{n}(3k+2) = \frac{3^n\,\Gamma\left(n+\frac{5}{3}\right)}{\Gamma\left(\frac{5}{3}\right)}\tag{1}$$ and the value of the RHS at $n=0$ equals $1$. It follows that the whole series can be represented as

$$S=\sum_{n\geq 0}\frac{1}{6^{n+1}(n+1)!}\cdot\frac{3^n\,\Gamma\left(n+\frac{5}{3}\right)}{\Gamma\left(\frac{5}{3}\right)}=\sum_{n\geq 0}\frac{1}{6\cdot 2^n}\cdot\frac{\Gamma\left(n+\frac{5}{3}\right)}{\Gamma\left(\frac{5}{3}\right)\Gamma(n+2)}\tag{2}$$ or, by using Euler's beta function and the reflection formula for the $\Gamma$ function, as $$\sum_{n\geq 0}\frac{1}{6\,\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{5}{3}\right)}\cdot\frac{B\left(n+\frac{5}{3},\frac{1}{3}\right)}{2^n}=\frac{\sqrt{3}}{8\pi}\sum_{n\geq 0}\int_{0}^{1}\frac{x^{n+2/3}(1-x)^{-2/3}}{2^n}\,dx \tag{3}$$ from which: $$S = \frac{\sqrt{3}}{4\pi}\int_{0}^{1}\frac{x^{2/3}}{(1-x)^{2/3}(2-x)}\,dx \stackrel{\frac{x}{1-x}\mapsto z}{=}\frac{\sqrt{3}}{4\pi}\int_{0}^{+\infty}\frac{z^{2/3}}{(1+z)(2+z)}\tag{4}$$ and: $$S = \frac{\sqrt{3}}{4\pi}\left(-\int_{0}^{+\infty}\frac{z^{2/3}}{z(1+z)}\,dz+2\int_{0}^{+\infty}\frac{z^{2/3}}{z(2+z)}\,dz\right).\tag{5}$$ Euler's beta function then leads to: $$S = \color{blue}{\frac{1}{\sqrt[3]{2}}-\frac{1}{2}} \approx 0.2937.\tag{6}$$

• please explain me how can we write $\displaystyle \prod_{k=1}^{n}(3k+2) = \frac{3^n\,\Gamma\left(n+\frac{5}{3}\right)}{\Gamma\left(\frac{5}{3}\right)}$. – DXT Sep 15 '18 at 7:21
• @DurgeshTiwari: by just exploiting the main functional identity for the $\Gamma$ function: $$\Gamma(n+5/3)=(n+2/3)\Gamma(n+2/3)=(n+2/3)(n-1/3)\Gamma(n-1/3)=\ldots$$ – Jack D'Aurizio Sep 15 '18 at 16:33
