# Equivalent Sums

Calu

## Homework Statement

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.

## Homework Equations

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.

Last edited:

Homework Helper
Gold Member

## Homework Statement

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.

## Homework Equations

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.

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

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

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?