Evaluation of a series (possibly related to Binomial Theorem)

Question

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.

Answer 1

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.

Define pr[n_] := Product[(3 i - 1)/(6 i), {i, 1, n}] Compute Sum[pr[n], {n, 0, [Infinity]}] Result: 2^(2/3)

Answer 2

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.