0

I need to calculate the sum :

$\sum_{k=1}^{n}\frac{1^2\space5^2\space...\space(4k-3)^2}{3^2\space7^2\space...\space(4k-1)^2}$

I am thinking of writing it as a difference of sums so when I do the difference some terms will reduce and we will get the first term from one and the last from the other one. Can you help me find out how to write it?

Ghost
  • 1,105
  • 2
    Is the sum supposed to be independent of $k$? – cansomeonehelpmeout Nov 03 '18 at 19:07
  • My bad. I edited it. – Ghost Nov 03 '18 at 19:08
  • 2
    Before getting to the how, may I ask you the why? If the purpose is to prove that the series over $k\geq 1$ is divergent, this has already been asked and answered on MSE, a few days ago. – Jack D'Aurizio Nov 03 '18 at 19:38
  • I doubt there's a closed form for the integral. $\sum _{k=1}^n \frac{\prod _{j=1}^k (4 j-3)^2}{\prod _{j=1}^k (4 j-1)^2}=\sum _{k=1}^n \frac{\Gamma \left(\frac{3}{4}\right)^2 \Gamma \left(\frac{1}{4}+k\right)^2}{\Gamma \left(\frac{1}{4}\right)^2 \Gamma \left(\frac{3}{4}+k\right)^2}=\int_0^1 \frac{\Gamma \left(\frac{3}{4}\right)^2 \left(2 \sqrt[4]{x} \left(-1+x^n\right) K(1-x)\right)}{\Gamma \left(\frac{1}{4}\right)^2 (\pi (-1+x))} , dx$ where $K$ is complete elliptic integral of the first kind. – Mariusz Iwaniuk Nov 03 '18 at 20:14
  • @JackD'Aurizio I just want to know how to calculate sums like this. – Ghost Nov 03 '18 at 21:36
  • In such a case, just have a look at the Maclaurin series of $K(x)$ (the complete elliptic integral of the first kind) on Wikipedia or Mathworld. – Jack D'Aurizio Nov 03 '18 at 21:37

0 Answers0