<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] 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:

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

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