binbagsss
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Homework Statement
I want to show that ## \sum\limits_{n=1}^{\infty} log (1-q^n) = -\sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty} \frac{q^{n.m}}{m} ##, where ##q^{n}=e^{2\pi i n t} ## , ##t## [1] a complex number in the upper plane.
Only that ## e^{x} = \sum\limits_{m=0}^{\infty} \frac{x^{m}}{m!}## [2]
I can see that both series start from ##n,m=1## in the RHS of [1] , so if I use [2] starting from ##m=1## will result in the '1' cancelling as needed i.e. I have ## \sum\limits_{n=1}^{\infty} log (1-q^n) = \sum\limits_{n=1}^{\infty} log (\sum\limits_{m=1}^{\infty} \frac{(2\pi int)^{m}}{m!})##
I don't really no where to go now, I don't see how you can get another expansion in terms of ##q^{m}## from this to give the required ##q^{n.m}##, I can see there's a minus sign too but I'm struggling to use this as a clue as well.
Many thanks in advance.
I want to show that ## \sum\limits_{n=1}^{\infty} log (1-q^n) = -\sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty} \frac{q^{n.m}}{m} ##, where ##q^{n}=e^{2\pi i n t} ## , ##t## [1] a complex number in the upper plane.
Homework Equations
Only that ## e^{x} = \sum\limits_{m=0}^{\infty} \frac{x^{m}}{m!}## [2]
The Attempt at a Solution
I can see that both series start from ##n,m=1## in the RHS of [1] , so if I use [2] starting from ##m=1## will result in the '1' cancelling as needed i.e. I have ## \sum\limits_{n=1}^{\infty} log (1-q^n) = \sum\limits_{n=1}^{\infty} log (\sum\limits_{m=1}^{\infty} \frac{(2\pi int)^{m}}{m!})##
I don't really no where to go now, I don't see how you can get another expansion in terms of ##q^{m}## from this to give the required ##q^{n.m}##, I can see there's a minus sign too but I'm struggling to use this as a clue as well.
Many thanks in advance.