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