Prove a series is the same as another

Question

How do I get from this series: $$\sum_{n=-\infty}^{\infty} (-1)^n e^{-2n^2\lambda^2}$$ to this series: $$1-2\sum_{n=1}^{\infty} (-1)^{n-1} e^{-2n^2\lambda^2}$$

I have proven numerically in python that they are equal, but I can't get to the answer analytically

I know the series is symmetrical so I change the indexes as $2\sum_{n=0}^{\infty} (-1)^{n} e^{-2n^2\lambda^2}$

Then I change the power to change signs as: $-2\sum_{n=0}^{\infty} (-1)^{n-1} e^{-2n^2\lambda^2}$

Then get the first index out so I can start the series in n=1, getting: $-2-2\sum_{n=1}^{\infty} (-1)^{n-1} e^{-2n^2\lambda^2}$

But this is not the same, what step am I missing?

Answer 1

$\sum_{n=-\infty}^{\infty} (-1)^n e^{-2n^2\lambda^2}$ assuming this converges

$= \sum_{n=-\infty}^{1} (-1)^n e^{-2n^2\lambda^2} + ((-1)^0 e^{-2(0)^2\lambda^2})+\sum_{n=1}^{\infty} (-1)^n e^{-2n^2\lambda^2}$ assuming the sums converge

$= \sum_{n=1}^{\infty} (-1)^{-n} e^{-2(-n)^2\lambda^2} + 1+\sum_{n=1}^{\infty} (-1)^n e^{-2n^2\lambda^2}$

$= 1 + \sum_{n=1}^{\infty} [ (-1)^{-n} e^{-2(-n)^2\lambda^2}+ (-1)^n e^{-2n^2\lambda^2}]$

$= 1 + \sum_{n=1}^{\infty}[ (-1)^{n} e^{-2n^2\lambda^2}+ (-1)^n e^{-2n^2\lambda^2}]$

$= 1 + \sum_{n=1}^{\infty}2 (-1)^{n} e^{-2n^2\lambda^2}$

$= 1 + \sum_{n=1}^{\infty}\frac 2{-1} (-1)^{n-1} e^{-2n^2\lambda^2}$

$=1 + \frac 2{-1}\sum_{n=1}^{\infty} (-1)^{n-1} e^{-2n^2\lambda^2}$

$=1 - 2\sum_{n=1}^{\infty}\frac (-1)^{n-1} e^{-2n^2\lambda^2}$

Answer 2

Then get the first index out so I can start the series in n=1, getting: $-2-2\sum_{n=1}^{\infty} (-1)^{n-1} e^{-2n^2\lambda^2}$

Substituting in zero gives you $-2(-1)^{0-1}e^{-2\lambda^2} = -2(-1)=2$. So one of your errors was having a negative sign for the first term.

I know the series is symmetrical so I change the indexes as $2\sum_{n=0}^{\infty} (-1)^{n} e^{-2n^2\lambda^2}$

All the negative indices result in the same value as for the corresponding positive index, so you can indeed remove the negative indices and double the values for the positive indices. But your error is failing to notice that there is an index that is neither positive nor negative: zero is a special case that needs to be handled separately. Your doubling would be valid if there were both $+0$ and $-0$, but since there is only one zero, you are double counting.

So one of your errors turned the $1$ into a $2$, and your other error turned the $2$ into a $-2$.