Here I have this sum, $\displaystyle \sum_{n=1}^\infty \displaystyle \prod_{k=1}^n \dfrac{2k+1}{4k}$.

I have no idea how to sum this up. Any help would be appreciated!


closed as off-topic by Namaste, Trevor Gunn, Jack D'Aurizio Sep 17 '18 at 16:21

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Trevor Gunn, Jack D'Aurizio
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Start by simplifying the summand by writing the product in terms of factorials and try to notice the connection of the summand to ${-3/2 \choose n}$. Then consider the power-series of $(1 - 4 x)^{-3/2} - 1$. $\endgroup$ – Winther Sep 17 '18 at 15:07
  • $\begingroup$ Please avoid "no-clue" questions. They do not make good quality questions and are likely to be removed. $\endgroup$ – Trevor Gunn Sep 17 '18 at 15:23

Is this true: $$\prod_{k=1}^n\frac{2k+1}{4k}=\frac{\prod_{k=1}^n2k+1}{\prod_{k=1}^n4k}=\frac{\prod_{k=1}^n2k+1}{4^n\prod_{k=1}^nk}=\frac{\Gamma(2n+1)}{4^n\Gamma(n)}$$

EDIT as pointed out by another user: $$\frac{\prod_{k=1}^n2k+1}{4^n\prod_{k=1}^nk}=\frac{(2n+1)!!}{4^nn!}$$

  • $\begingroup$ I'm not yet familiar with Gamma function. $\endgroup$ – user591656 Sep 17 '18 at 15:11
  • $\begingroup$ I am familiar with it but not in this context, so I need someone else to verify that this is true $\endgroup$ – Henry Lee Sep 17 '18 at 15:12
  • 2
    $\begingroup$ $\prod_{k=1}^n (2k + 1) = (2n + 1)(2n - 1) \cdots 3 \cdot 1 = (2n + 1)!! = \frac{(2n + 2)!}{2^{n + 1}(n + 1)!}$ and $\prod_{k = 1}^n k = n! = \Gamma(n + 1)$. $\endgroup$ – Trevor Gunn Sep 17 '18 at 15:12

Not the answer you're looking for? Browse other questions tagged or ask your own question.