# Equivalent Sums

1. Dec 10, 2014

### Calu

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

I honestly have no idea what may be relevant, other than what is above.

3. 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.

Last edited: Dec 10, 2014
2. Dec 10, 2014

### LCKurtz

Make the change of variable $n = 2m$ in the first sum and see what you get.

3. Dec 10, 2014

### Calu

Right I see, this would give me what I have in the second sum, obviously with m substituted for n. So can this change of variable be used as I'm summing over twice as many integers?

4. Dec 10, 2014

### LCKurtz

All the integers is not "twice as many integers". They are both infinite sets. But you just need to note that all even integers are given by $2m$ as $m = 1,2,3....$. The index goes through all the natural numbers but $2m$ gives all the evens.

5. Dec 10, 2014

### Calu

I see now, thanks for your help.