# Finding the sum of the infinite series whose general term is not easy to visualize: $\frac16+\frac5{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\cdots$ [duplicate]

I am to find out the sum of infinite series:- $$\frac{1}{6}+\frac{5}{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+...............$$ I can not figure out the general term of this series. It is looking like a power series as follows:- $$\frac{1}{6}+\frac{5}{6^2\cdot2!}+\frac{5\cdot8}{6^3\cdot3!}+\frac{5\cdot8\cdot11}{6^4\cdot4!}+.....$$ So how to solve it and is there any easy way to find out the general term of such type of series?

• Just to be clear, when you write $6.12$, you mean multiplication $6$ times $12$, yeah? – Mike Pierce Jan 5 '16 at 3:37
• Apart from the first term which doesn't follow the pattern, it appears that each additional term is of the form $\frac{1}{6}\prod\limits_{n=0}^k\frac{3n+5}{6n+12}$. The missing fifth entry by this pattern would be $\frac{5\cdot 8\cdot 11\cdot 14}{6\cdot 12\cdot 18\cdot 24\cdot 30}$, etc... Without knowing exactly where the sequence comes from, it is not guaranteed however that this actually is the pattern, it is just a guess. If that is the case, then it will converge slightly faster than a geometric series with $r=\frac{1}{2}$, so it should be able to be approximated. – JMoravitz Jan 5 '16 at 3:39
• yes.....if there is any confusion i may edit it – Mayank Deora Jan 5 '16 at 3:39
• @JMoravitz: The first term does fit the pattern: it's $1/6$ times the empty product. – Rahul Jan 5 '16 at 4:27
• this is not an answerable question – Arjang Jan 5 '16 at 7:43

Notice \begin{align} & \frac{1}{6} + \frac{5}{6\cdot 12} + \frac{5\cdot 8}{6\cdot 12\cdot 18} + \cdots\\ = & \frac12 \left[ \frac{3-1}{6} + \frac{(3-1)(6-1)}{6^2 \cdot 2!} + \frac{(3-1)(6-1)(9-1)}{6^33!} + \cdots\right]\\ = & \frac12 \left[ \frac{\frac23}{2} + \frac{\frac23(\frac23+1)}{2^2\cdot 2!} + \frac{\frac23(\frac23+1)(\frac23+2)}{2^3\cdot 3!} + \cdots \right] \end{align} The sum we want has the form $\displaystyle\;\frac12 \sum_{k=1}^\infty \frac{\prod_{\ell=0}^{k-1}(\frac23 + \ell)}{2^k k!}$. Compare this with the expansion

$$\frac{1}{(1-z)^\alpha} = \sum_{k=0}^\infty \frac{(\alpha)_k}{k!} z^k \quad\text{ where }\quad (\alpha)_k = \prod\limits_{\ell=0}^{k-1}( \alpha + \ell )$$

We find the desired sum equals to $$\frac12 \left[\frac{1}{\left(1-\frac12\right)^{2/3}} - 1\right] = \frac12\left( 2^{2/3} - 1 \right) \approx 0.293700525984...$$

Notes

After I finish this answer, I have a déjà vu feeling that I have seen this before (more than once).
Following are some similar questions I can locate, I'm sure there are more...

• how come I have a déjà vu feeling that I have seen this before??? – achille hui Jan 5 '16 at 4:19
• It is a question from NBHM 2011 sample paper ,so it is quite possible you may have seen it before – Mayank Deora Jan 5 '16 at 5:06
• The general term can be made anything we want it to be, this is same as given a finite sequence trying to guess the next term. without the explicit formula for the general term or some recursive way of defining the terms it is just waste of time to guess the general fprm. – Arjang Jan 5 '16 at 7:42

Let us consider $$\Sigma=\frac{1}{6}+\frac{5}{6\times 12}+\frac{5\times8}{6\times12\times18}+\frac{5\times8\times11}{6\times12\times18\times24}+\cdots$$ and let us rewrite it as $$\Sigma=\frac{1}{6}+\frac 16\left(\frac{5}{ 12}+\frac{5\times8}{12\times18}+\frac{5\times8\times11}{12\times18\times24}+\cdots\right)=\frac{1}{6}+\frac 16 \sum_{n=0}^\infty S_n$$ using $$S_n=\frac{\prod_{i=0}^n(5+3i)}{\prod_{i=0}^n(12+6i)}$$ Using the properties of the gamma function, we have $$\prod_{i=0}^n(5+3i)=\frac{5\ 3^n \Gamma \left(n+\frac{8}{3}\right)}{\Gamma \left(\frac{8}{3}\right)}$$ $$\prod_{i=0}^n(12+6i)=6^{n+1} \Gamma (n+3)$$ which make $$S_n=\frac{5\ 2^{-n-1} \Gamma \left(n+\frac{8}{3}\right)}{3 \Gamma \left(\frac{8}{3}\right) \Gamma (n+3)}$$ $$\sum_{n=0}^\infty S_n=\frac{10 \left(3\ 2^{2/3}-4\right) \Gamma \left(\frac{2}{3}\right)}{9 \Gamma \left(\frac{8}{3}\right)}=3\ 2^{2/3}-4$$ $$\Sigma=\frac{1}{\sqrt[3]{2}}-\frac{1}{2}$$

This is $\frac{1}{2} \left(_2F_1(2/3,1;1;\frac{1}{2}) - 1\right)$ where $_2F_1$ is the hypergeometric function. For these parameters this evaluates to $\frac{1}{2} \left(_2F_1(2/3,1;1;\frac{z}{2}) - 1\right) = \frac{1}{2} \left( (1 - \frac{z}{2})^{-2/3} - 1\right)$.

It seems like (without knowing the exact form we can not proceed) the double of $n$th term of your series is given by $$u_n=\dfrac{2.5.8.11\cdots(3n-1)}{6.12.18.24\cdots(6n)}.$$ By ratio test this series is convergent and we can find the exact formula for $n$th partial summation $\sum^{n}_{r=1} u_r$ by observing $$(6n+6)u_{n+1}=u_n(3n+2)\forall n\in\mathbb{N}.$$ Find a function $f$ with the property $$f(r+1)=f(r)\left(\dfrac{3r+2}{6r+6}\right)$$ (in terms of $u_r$) and $$u_r=f(r)-f(r+1).$$ Then Telescope to find $\sum^{n}_{r=1} u_r$.