I have that, for n ∈ ℕ, (-6/π) ∑n even (cos(nx))/(n2-9) is equivalent to (-6/π) ∑∞n=1 (cos(2nx))/(4n2-9). I don't understand how the two sums are equivalent to each other.
I honestly have no idea what may be relevant, other than what is above.
The Attempt at a Solution
I'm making a guess that the factor of 2 within the cosine function in the second sum allows the sum to be the same as summing over the even natural numbers. However I'm not sure that's correct, and I'm still unsure where the factor of 4 comes from in front of the n2 in the denominator, other than it is 22.