In summary, using the fact that the sum of series 1/n^2 is (pi)^2/6, we can find the value of cos(2npi/3)/(n^2) by splitting the series into two parts and evaluating them separately. First, we can calculate the sum of 1/(3n)^2 which gives us pi^2/54. Then, we can use the identity \sum_{n=1}^{+\infty}{\frac{1}{n^2}}=\frac{\pi^2}{6} to find the sum of 1/(3n+1)^2 and 1/(3n+2)^2. By substituting these values into the equation \
#36
andrey21
476
0
Rite so what your saying is:
Pi^2/6 = Pi^2/54 + SUM 1/(3n+1)^2 + SUM 1/(3n+2)^2
Well, in post 14, we were asked to calculate those two sums. I first tried to evaluate them separately, but that didn't work. So then I came up with that solution. I guess it's a bit experience from my part. The more problems you solve, the more tricks you know. You also know the trick now
#42
andrey21
476
0
Haha ok thanks again micromass
#43
andrey21
476
0
I just finished my calculations and ended up with value of -3Pi^2/54 for sum of series. Is this correct?? Thanks in advance