# Sum of given infinite series

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?

• 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. – Simply Beautiful Art Jan 16 '17 at 16:00
• The answer is $1/3$, apparently. So, it should telescope somehow – Yuriy S Jan 16 '17 at 16:05
• @YuriyS How "apparently"? – DonAntonio Jan 16 '17 at 16:08
• @DonAntonio, Mathematica did it for me. Sum[k Product[1/(3 n + 1), {n, 1, k}], {k, 1, Infinity}] – Yuriy S Jan 16 '17 at 16:09
• @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. – Simply Beautiful Art Jan 16 '17 at 16:09

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.

• Nice observation – lab bhattacharjee Jan 17 '17 at 15:06
• @labbhattacharjee :-) Thanks! :D – Simply Beautiful Art Jan 17 '17 at 15:07

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.

• But... I want to make it come out prettier. – Simply Beautiful Art Jan 16 '17 at 16:23
• @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? – Daniel Fischer Jan 16 '17 at 16:30
• @DanielFischer :D Yup, much prettier. But I meant I wanted telescoping stuffs... Hm... thanks for the tip. – Simply Beautiful Art Jan 16 '17 at 17:30