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April 18, 2015

April 18, 2015

Posted by **COFFEE** on Saturday, July 28, 2007 at 12:58pm.

Infinity of the summation n=0: [(-1)^n pi^(2n)] / [6^(2n) (2n)!]

this is my work:

[(-1^0) pi^(2*0)] / [6^(2*0) (2*0)!] + [(-1^1) pi^(2*1)] / [6^(2*1) (2*1)!] + [(-1^2) pi^(2*2)] / [6^(2*2) (2*2)!] + [(-1^3) pi^(2*3)] / [6^(2*3) (2*3)!]

-1 + -0.13707783 + -0.00313172 + -0.0000286 + -0.00000014

sum of the series = -1.14023829

You got the signs wrong. The answer is 1/2 sqrt[3]

thank you!!!

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