Calculating cos(2nπ/3)/(n²) using the Sum of Series 1/n²

[micromass]

Yes! So that's the first sum. Now try to evaluate

$$\sum_{n=1}^{+\infty}{\frac{1}{(2n+1)^2}}+\sum_{n=1}^{+\infty}{\frac{1}{(2n+2)^2}}$$

 

[andrey21]

Ye ok is it (2n+2)^2 or (3n+2)^2? Just in the first page it was 3 not 2 before the n.

 

[micromass]

Ow, my apologies, it is 3 instead of 2...

 

[andrey21]

Ok so from the two sums I get:

1/(9n^2+6n+1)

and

1/(9n^2 +12n+4)

 

[micromass]

Hmm, you won't be able to calculate those sums separately...
You'll have to use the following:

$$\sum_{n=1}^{+\infty}{\frac{1}{n^2}}=\sum_{n=1}^{+\infty}{\frac{1}{(3n)^2}}+\sum_{n=1}^{+\infty}{\frac{1}{(3n+1)^2}}+\sum_{n=1}^{+\infty}{\frac{1}{(3n+2)^2}}$$

You know two of the above series...

 

[andrey21]

Rite so what your saying is:

Pi^2/6 = Pi^2/54 + SUM 1/(3n+1)^2 + SUM 1/(3n+2)^2

 

[micromass]

Yes, so you've found the values to all the series!

 

[andrey21]

Ah ok so:

Pi^2/54 - 1/2(Pi^2/6 - Pi^2/54) = Sum of series

 

[micromass]

Yes! Very good!

 

[andrey21]

Thanks Micromas you have been great help. Out if curiosity how did you obtain the equation in post 35 ??

 

[micromass]

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

 

[andrey21]

Haha ok thanks again micromass

 

[andrey21]

I just finished my calculations and ended up with value of -3Pi^2/54 for sum of series. Is this correct?? Thanks in advance

 

[micromass]

Yes, I believe that is correct!