Sum of product result given by wolfram: $\sum_{k=1}^\infty\prod_{i=1}^k\frac{3i-1}{4i}=2(2)^\frac{1}{3}-1$

Question

So according to Wolfram Alpha, the following result holds: $$\sum_{k=1}^\infty\prod_{i=1}^k\frac{3i-1}{4i}=2(2)^\frac{1}{3}-1$$

Could someone explain to me how Wolfram Alpha gets this or how to calculate a sum of products like this one?

Answer 1

We use the generalized binomial theorem for general exponents $r\in \mathbb{C}$ with $x,y\in \mathbb{R}$ such that $|x|>|y|$: $$(x+y)^r=\sum_{k=0}^\infty {r \choose k} x^{r-k} y^k$$ with $$\binom{r}{k}=\frac{r(r-1) \cdots (r-k+1)}{k!}$$

We then have that with $m:=1/3$ and $r:=-2/3$: $$\begin{align} \sum_{k=1}^{\infty} \prod_{i=1}^k \frac{3i-1}{4i}&=\sum_{k=1}^{\infty} \prod_{i=1}^{k} \left[\frac{3}{4}-\frac{1}{4i}\right]\\&=\sum_{k=1}^{\infty} \left(\frac{3}{4}\right)^k\prod_{i=1}^{k} \left[1-\frac{m}{i}\right]\\&=\sum_{k=1}^{\infty} \left(\frac{3}{4}\right)^k (1-m)(1-m/2)\cdots (1-m/k)\\&=\sum_{k=1}^{\infty} \left(-\frac{3}{4}\right)^k\cdot \frac{(m-1)(m-2)\cdots (m-k)}{k!}\\&=\sum_{k=1}^{\infty} \left(-\frac{3}{4}\right)^k\cdot \frac{r(r-1)\cdots (r-k+1)}{k!}\\&=\sum_{k=0}^{\infty} \left[\frac{r(r-1)\cdots (r-k+1)}{k!}\cdot 1^{r-k}\cdot (-3/4)^k\right]-1\\&=(1-3/4)^{-2/3}-1\\&=2\cdot 2^{1/3}-1\end{align}$$ which is what we wanted to prove.