Friday

March 6, 2015

March 6, 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!!!

**Answer this Question**

**Related Questions**

Calculus - Show that the following series is absolutely convergent: Summation ...

Calculus 2 - In the following series x is a real number. In each case, use the ...

CALC 2 - In the following series x is a real number. In each case, use the ratio...

CALC 2 pls help!! - In the following series x is a real number. In each case, ...

Calculus III - Which of the following series are geometric series? Find the sum ...

calculus - ratio test - infinity of the summation n=1: (e^n)/(n!) [using the ...

calculus - ratio test - Posted by COFFEE on Sunday, July 29, 2007 at 6:32pm. ...

Calculus - ratio test - infinity of the summation n=1: (e^n)/(n!) [using the ...

Calculus - Does the series (1+sin(n))/(10^n) from summation 0 to positive ...

Calculus - For what values of p>0 does the series Riemann Sum [n=1 to ...