find the sum of the series?
sigma notation (( (-1)^n * pi^2n)) / (6^2n * (2n)!) as n=0 goes to ininity
sigma notation (( (-1)^n * pi^2n)) / (6^2n * (2n)!) as n=0 goes to ininity
robust
Favorite Answer
The maclaurin series for cos x is
1 - x^2/2! + x^4/4! - x^6/6!....
If we let x = pi/6 in the above expansion we recognize that your series is the series of cospi/6
√3/2
latassa
Find The Sum Of Series
Anonymous
It's a cos expansion evaluated at Pi/6 =>
=cos(Pi/6)=Sqrt(3)/2
kevin b
(-1)^n (Pi^2n)/(6^2n*(2n)!) use ratio test
(Pi^2(n+1)/6^2n*(2(n+1)!))*(6^2n*(2n)!)/(Pi^2n)
Pi^2*(2n)!/(2n+2)(2n+1)(2n!)
Pi^2/(2n+2)(2n+1)=Pi^2/INFINITY
=0