# Equivalent Sums

## 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:

Related Precalculus Mathematics Homework Help News on Phys.org
LCKurtz
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.

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?

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.
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.