Question

Infinite product =)

Answer 1

\[\Large \frac{2^5 4^4 6^4 8^4 ... }{3^5 5^4 7^4 9^4... }\] Only the first two numbers have an extra power of 5, after that it's all even powers of 4 on top and odd powers of 4 on bottom. Good luck, this has a nice answer. =)

Answer 2

\[\frac{e^2}{4}\]?

Answer 3

Surprisingly close, but not quite. =P

Answer 4

lol i remember a guy hear a few years ago (i think it was alchemista) asked questions like this so one time i just wrote \(\frac{e^2}{4}\) and he wanted to know how i got it so fast

Answer 5

*here

Answer 6

Hahaha, inkyvoid was like literally a step or two away from the answer. I guess you could keep guessing @satellite73 since you were pretty close too hahaha XD

Answer 7

0 ?

Answer 8

wait it's not the same as the answer to the basel problem is it?

Answer 9

You got it @inkyvoyd

Answer 10

\(\frac{\pi^2}{16}\)

Answer 11

Solution: \[\Large \frac{2^5 4^4 6^4 8^4 ... }{3^5 5^4 7^4 9^4... }\] factor out 2/3 and a 2 from the exponents:
\[\Large \frac{2}{3} \left( \frac{2^2 4^2 6^2 8^2 ... }{3^2 5^2 7^2 9^2... } \right)^2 \] use pi/2 infinite sum, plug it in:

\[\Large \frac{2}{3} \left( \frac{ \pi }{2 } \right)^2 = \frac{ \pi ^2}{6} \]

Answer 12

http://arxiv.org/ftp/arxiv/papers/1005/1005.2712.pdf

Answer 13

im going through that link has something about why that infinite ratio equals pi/2

Answer 14

Kinda cool that we can relate an infinite product and infinite series through a funny thing of pi. Yeah it seems totally bizarre, how do you figure something like this out?

Answer 15

strange indeed
http://math.stackexchange.com/questions/700686/equality-between-an-infinite-product-and-an-infinite-series-how-can-i-reconcile

Answer 16

I think this is all because of gamma function
\[\dfrac{2^23^24^25^2\cdots}{3^25^27^29^2\cdots} = \prod_{k=0}^{\infty}
\dfrac{(2k+2)(2k+2)}{(2k+1)(2k+3)} = \prod_{k=0}^{\infty}\dfrac{(k+1)(k+1)}{(k+1/2)(k+3/2)}\]

next appealing to this thm
http://gyazo.com/b0a805580ffd10e924bbf7d92aa9cfcb

\[=\dfrac{\Gamma(1/2)\Gamma(3/2)}{\Gamma(1)^2} = \frac{1}{2}\Gamma(1/2)^2 = \frac{\pi}{2}\]