# Evaluation of a series (possibly related to Binomial Theorem)

I have the following series:

$$1 + \frac{2}{3}\cdot\frac{1}{2} + \frac{2\cdot5}{3\cdot6}\cdot\frac{1}{2^2} + \frac{2\cdot5\cdot8}{3\cdot6\cdot9}\cdot\frac{1}{2^3} + \ldots$$

I have to find the value of this series, and I have four options:
(A) $$2^{1/3}$$ (B) $$2^{2/3}$$ (C) $$3^{1/2}$$ (D) $$3^{3/2}$$

I can't seem to find a general term for this. I tried:

$$S = 1 + \frac{(1 - \frac{1}{3})}{1!}(\frac{1}{2}) + \frac{(1 - \frac{1}{3})(2 - \frac{1}{3})}{2!}(\frac{1}{2})^2 + \frac{(1 - \frac{1}{3})(2 - \frac{1}{3})(3 - \frac{1}{3})}{3!}(\frac{1}{2})^3 + \ldots$$

But this doesn't seem to get me anywhere.

Any help?

This maybe a telescopic series, because there was a similar question we solved in class which ended up being telescopic:

$$\frac{3}{2^3} + \frac{4}{2^4\cdot3} + \frac{5}{2^6\cdot3} + \frac{6}{2^7\cdot5} + \ldots$$

$$=\displaystyle\sum\limits_{r=1}^\infty\frac{r+2}{2^{r+1}r(r+1)}$$

$$=\displaystyle\sum \bigg(\frac{1}{2^r r} - \frac{1}{2^{r+1}(r+1)}\bigg) = \frac{1}{2}$$

$$P.S:$$ This problem was included in my set of questions for Binomial Theorem, which is why I thought it might be related to it.

• $$\sum_{n=0}^\infty\frac{(3n-1)!!!}{(3n)!!!}2^{-n}=?$$ And it's not related to Newton's binomial. Commented Nov 5, 2013 at 5:17
• @Lucian What does the triple $!$ mean? Is it a nested factorial? And I've edited my question too :) Commented Nov 5, 2013 at 6:02
• Just like the double factorial; only that it's triple. :-) Commented Nov 5, 2013 at 6:05
• It can't be telescopic, can it, since $\sqrt[3]4$ is not a rational quantity. Commented Nov 5, 2013 at 6:21

You're looking for $$1+\sum_{k=1}^\infty \frac{\prod_{i=1}^k (3i-1)}{3^k k!}2^{-k}$$ That looks an awful lot like a Maclaurin series to me. What sort of a function would give us a Maclaurin series like this? I started with $f(x)=x^{-2/3}$, since continued derivation should give coefficients similar to those seen in the sum. Obviously we don't want to find the series around $0$, so we'll center the function around 1: $f(x-1)=(x-1)^{-2/3}$. If you compute the Maclaurin series, you obtain $$1+\sum_{k=1}^\infty \frac{\prod_{i=1}^k(3i-1)}{3^k}\frac{x^k}{k!}=(x-1)^{-2/3}$$ So $(\frac{1}{2}-1)^{-2/3}=\sqrt[3]{4}$ is your sum.
• If you'd be so kind, could you tell me what Taylor-Maclaurin series yields the following result : $$1-5\cdot\left(\frac12\right)^3+9\cdot\left(\frac{1\cdot3}{2\cdot4}\right)^3- ... =\frac2\pi$$ Commented Nov 5, 2013 at 7:11
• I think I'd need to see more terms of that sequence, but it looks like $1+\sum (-1)^n(4n+1)(\frac{(2n)!}{4^n n!^2})^3$, which I think would correspond to some sort of root, since the series for sqrt(x+1) has a similar sort of pattern (without the cubes... those look problematic). Commented Nov 5, 2013 at 18:36
• It's from Ramanujan's Second Notebook. There aren't any more explicit terms, but we're obviously dealing with $$\sum_{n=0}^\infty(-1)^n\cdot(4n+1)\cdot\left[\frac{(2n-1)!!}{(2n)!!}\right]^3=\frac2\pi$$ Commented Nov 5, 2013 at 21:34
• That sounds like a question you may want to put on the forum, but the general idea of a Maclaurin series is $$f(x-n)=\sum_{k=0}^\infty f^{(k)}(n)\cdot\frac{x^k}{k!}$$ Where $f^{(k)}(n)$ is the $k^{\mathrm{th}}$ derivative of $f$ at $n$. If you have an infinite sum of the terms of a Maclaurin series, then you can just replace it with the original function and evaluate that function. Commented Nov 10, 2013 at 8:15
According to Mathematica, the sum is $2^{2/3}$. I don't know how to ask Mathematica to explain its method for deriving that result.