lets see how many can solve this?

Question

i made this question yesterday. the first correct answer will win.

lim(n→∞)

n
∏ [2^{ ( (-1)^r)·(π^(2r-1) ) / ( (2r-1)!) } ] = ?
r=1

12 answers, and no one right yet.
well, you can't solve that with a calculator, you need a brain with paper and pencil/pen.
i'm extending the expiration to 7 days.

u got it Cody A, right on the 13th answer.

i know the solution, but i want it. Cody A, 1 of the answers is correct. but i want the correct 1 and the solution.

i'm really sorry, but i didn't see avocaronico's answer. it was him who was the first.
the infinite product becomes

2^(-π+π^3/3!-π^5/5!+π^7/7!...∞)
=2^(-sin π)
=2^(-0)
=1

Answer 1

The exponent expands into the series expansion for Sin[Pi], which is 0. Since 2^0 = 1, the answer is 1.

Answer 2

36

Answer 3

So you got the result? What are you waiting for? Give as many people 2 points as much as you can?
(BTW thanx for helping me to get 2 easy points)

Answer 4

Are you sure it even converges?? That -1^r in the exponent makes me wonder........

Anyway, if it converges to anything, it has to be 0 (because of the alternating sign of each succesive term ☺)


Doug

Answer 5

It's undefined. The factorial grows faster than the exponent in the numerator. Therefore the fraction goes to 1/0.

Edit -----------------------------------------------
oh, I thought the · was a -. Silly unicode math nonsense.

Answer 6

no need of sorry if u telling me sorry
its my fault that i didnt give u the solution
sorry

Answer 7

Strangely enough, the answer is "Tuna"

Answer 8

HEE HEE ITS A SPOON!!!


it talks to me...

WOOOOOOO!!!!

CrAZY AttACk MONKeyS!

DO YOU SMELL the monkey??

I am a Squirrel...Cheep cheep.

Answer 9

pineapple

Answer 10

8.824977827