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Find the sum of infinite series

$$\frac{1}{4}+\frac{2}{4 \cdot 7}+\frac{3}{4 \cdot 7 \cdot 10}+\frac{4}{4 \cdot 7 \cdot 10 \cdot 13 }+....$$

Generally I do these questions by finding sum of $n$ terms and then putting $ \lim{n \to \infty}$ but here I am not able to find sum of $n$ terms. Could some suggest as how to proceed?

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    The mild hint is to perform PFD on a ratio of Gamma functions to construct a telescoping series, if anyone's able to see it. – 2017-01-16
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    The answer is $1/3$, apparently. So, it should telescope somehow – 2017-01-16
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    @YuriyS How "apparently"? – 2017-01-16
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    @DonAntonio, Mathematica did it for me. `Sum[k Product[1/(3 n + 1), {n, 1, k}], {k, 1, Infinity}]` – 2017-01-16
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    @DonAntonio As I said, a ratio of gamma functions. We can probably see why the gamma functions should be here, but the challenge is to make it telescope. – 2017-01-16
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    @SimpleArt Perhaps, but I've no idea how to tackle this that way... – 2017-01-16
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    @DonAntonio Me neither, I haven't done a problem like this that way yet XD I only see it, but its not within my reach... – 2017-01-16
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    Just a comment. If we consider the function $$f(x)=\sum_{k=1}^\infty \frac{x^k}{\prod_{n=1}^k (3n+1)}$$ then we have $$S=f'(1)$$ – 2017-01-16
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    @YuriyS Ask mathematica for the finite sum and it should make the telescope more obvious. – 2017-01-16
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    @SimpleArt here you have it http://prntscr.com/dwdh6h – 2017-01-16
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    @SimpleArt, the result is $$\frac{3^{-N-2} \Gamma \left(\frac{4}{3}\right) \left(-3 N \Gamma \left(\frac{7}{3}\right)+4\ 3^N \Gamma \left(\frac{1}{3} (3 N+7)\right)-4 \Gamma \left(\frac{7}{3}\right)\right)}{\Gamma \left(\frac{7}{3}\right) \Gamma \left(\frac{1}{3} (3 N+7)\right)}$$ – 2017-01-16
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    @YuriyS Haha, never mind then :P – 2017-01-16
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    It is the same as $\sum_{n=1}^{\infty} (n)(\frac{1}{3})^{n+2} \frac{\Gamma (\frac{1}{3})}{\Gamma (n+2+\frac{1}{3})}$ – 2017-01-16
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    @AhmedS.Attaalla Are you hinting at the beta function? – 2017-01-16
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    It looks pretty similar, but not sure how it will help. Just thought to leave that there just in case someone more knowledgeable would find it useful @SimpleArt – 2017-01-16

2 Answers 2

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Notice that

$$\frac k{\prod_{m=1}^k(3m+1)}=\frac1{3\prod_{m = 1}^{k-1} (3m+1)}-\frac{1}{3\prod_{m = 1}^k (3m+1)}$$

Which gives us a telescoping series:$$S_N=\frac{1}{3} - \frac{1}{3\prod_{m = 1}^N (3m+1)}$$

which tends to $1/3$ as suspected.

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    Nice observation – 2017-01-17
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    @labbhattacharjee :-) Thanks! :D – 2017-01-17
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The partial sums, according to Maple, are $$-{\frac {2\,{3}^{1/2-N}\pi}{27\,\Gamma \left( 4/3+N \right) \Gamma \left( 2/3 \right) }}+\frac{1}{3} $$ It should be possible to prove that by induction.

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    But... I want to make it come out prettier. – 2017-01-16
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    @SimpleArt $$\sum_{k = 1}^N \frac{k}{\prod_{m = 1}^k (3m+1)} = \frac{1}{3} - \frac{1}{3\prod_{m = 1}^N (3m+1)}$$ pretty enough? – 2017-01-16
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    @DanielFischer :D Yup, much prettier. But I meant I wanted telescoping stuffs... Hm... thanks for the tip. – 2017-01-16