Sum of $1+\frac{1}{2}+\frac{1\cdot2}{2\cdot5}+\frac{1\cdot 2\cdot 3}{2\cdot 5\cdot 8}+\cdots$ I am trying to find out the sum of this
$$1+\frac{1}{2}+\frac{1\cdot2}{2\cdot5}+\frac{1\cdot 2\cdot 3}{2\cdot 5\cdot 8}+\frac{1\cdot 2\cdot 3\cdot 4}{2\cdot 5\cdot 8\cdot 11}+\cdots$$.
I tried with binomial theorem with rational index. But in vain. What shall I do to solve it?
If it has been solved already, kindly provide me the link. I am unable to get that. 
Thanks for your help
 A: Since $$S=1+\sum_{k=0}^{+\infty}\prod_{j=0}^k\frac{j+1}{3j+2}=1+\sum_{k=1}^{+\infty}k\, 3^{-k} B(2/3,k),$$ 
where $B(2/3,k)$ is the Euler Beta function:
$$ B(2/3,k)=\int_{0}^{1}x^{-1/3}(1-x)^{k-1}$$
we have: $$S=1+3\int_{0}^{1}\frac{dx}{(1-x)^{1/3}(3-x)^2}.$$
Now the last integral can be computed explicitly, and leads to:

$$S=\frac{3}{2}-2^{-4/3}\sqrt{3}\operatorname{arccot}\left(\frac{1-2^{4/3}}{\sqrt{3}}\right)+2^{-7/3}\log\left(4+2^{4/3}-2^{5/3}\right)-2^{-4/3}\log\left(2+2^{2/3}\right).$$

A: Here is an another approach that doesn't add anything much to Jack's answer but is at least mildly entertaining.  It does not use the $B$ function but instead takes a detour via a "simple" alternating series.  As stated in the comments the sum equals $${}_2F_1(1,1,\tfrac{2}{3};\tfrac{1}{3})$$ and using the transformation $${}_2F_1(\alpha, \beta, \gamma; z) = (1-z)^{-\alpha}{}_2F_1(\alpha, \gamma - \beta, \gamma; \tfrac{z}{z-1})$$ we get:
$$
\begin{eqnarray}
{}_2F_1(1,1,\tfrac{2}{3};\tfrac{1}{3}) &=& \tfrac{3}{2}{}_2F_1(1, -\tfrac{1}{3}, \tfrac{2}{3}; -\tfrac{1}{2})\\[1ex]
&=& \tfrac{3}{2} + \sum_{n=0}^{\infty}(-1)^n\frac{3}{2^{n+2}(3n+2)}\\
&=&\tfrac{3}{2} + 2^{-\frac{4}{3}}\int_0^{2^{-\frac{1}{3}}}\frac{3x}{1+x^3}dx\\[1ex]
&=&\tfrac{3}{2} + \tfrac{3}{2} \int_0^1 \frac{x}{2+x^3}dx\\[1ex]
\end{eqnarray}$$
