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

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:
$$
\begin{eqnarray}
{}_2F_1(1,1,\tfrac{2}{3};\tfrac{1}{3}) &=& \tfrac{3}{2}{}_2F_1(1, -\tfrac{1}{3}, \tfrac{2}{3}; -\tfrac{1}{2})\\[1ex]
&=& \tfrac{3}{2} + \sum_{n=0}^{\infty}(-1)^n\frac{3}{2^{n+2}(3n+2)}\\
&=&\tfrac{3}{2} + 2^{-\frac{4}{3}}\int_0^{2^{-\frac{1}{3}}}\frac{3x}{1+x^3}dx\\[1ex]
&=&\tfrac{3}{2} + \tfrac{3}{2} \int_0^1 \frac{x}{2+x^3}dx\\[1ex]
\end{eqnarray}$$
