1. ## Summation Identity

Let's try this again!

This is a post by Rmercadojr:

I've been trying to show that $\sum_{n=-\infty}^{\infty} q^{n(2n-1)} = \sum_{n=0}^{\infty} q^{n(n+1)/2}$ but I haven't been successful. I've managed to break the LHS into three pieces corresponding to the indices $-\infty$ to $-1$, $0$, and $1$ to $\infty$. Then I changed variables to get $1 + \sum_{n=1}^{\infty} q^{n(2n-1)} + \sum_{n=1}^{\infty} q^{n(2n+1)}$. I then added the two sums, but I can't find a way to manipulate it into the RHS.

I see how the LHS equals the RHS by expanding the terms, but I just haven't been able to show it algebraically. I found the identity $\sum_{n=0}^t f(2n) + \sum_{n=0}^t f(2n+1) = \sum_{n=0}^{2t+1} f(n)$. This seems like a very basic identity but I had never seen it. I haven't been able to find many examples of it being used either in books or online. Does anyone know a website where I can read more about it?

I'm sure that I can use this identity to solve my problem, but I haven't been able to. I will appreciate any help. Thank you.

2. ## Re: Summation Identity

Show (by induction on k)

$\displaystyle \sum _{n=-k}^k q^{n(2n-1)}=\sum _{n=0}^{2k} q^{n(n+1)/2}$

3. ## Re: Summation Identity

My question is more about the identity $\sum_{n=0}^t f(2n) + \sum_{n=0}^t f(2n+1) = \sum_{n=0}^{2t+1} f(n)$ and how to use it, its proof, or anything about it. Does it apply even when the indices are $\infty$? How do I apply it to my question? I'm sure that to solve my original question I just need a sort of trick or clever manipulation, not induction.

If it helps, what I am trying to prove is a special case of Ramanujan's generalized theta function, $\psi(q)$.

4. ## Re: Summation Identity

My question is more about the identity $\sum_{n=0}^t f(2n) + \sum_{n=0}^t f(2n+1) = \sum_{n=0}^{2t+1} f(n)$ and how to use it, its proof, or anything about it. Does it apply even when the indices are $\infty$? How do I apply it to my question? I'm sure that to solve my original question I just need a sort of trick or clever manipulation, not induction.
If it helps, what I am trying to prove is a special case of Ramanujan's generalized theta function, $\psi(q)$.
Let $\displaystyle f(n)=q^{n(n+1)/2}$
$\displaystyle f(-2n)=f(2n-1)=q^{n(2n-1)}$
$\displaystyle \sum _{n=0}^{2t+1} q^{n(n+1)/2}=\sum _{n=0}^{2t+1} f(n)=\sum _{n=0}^t f(2n)+\sum _{n=0}^t f(2n+1)=$
$\displaystyle \sum _{n=-t}^0 f(-2n)+\sum _{n=1}^{t+1} f(2n-1)=\sum _{n=-t}^{t+1} q^{n(2n-1)}$