Lambert series

1. Aug 22, 2006

lokofer

If we have the next Lambert series:

$$S(x)= \sum_{n=0}^{\infty} \frac{ a(n) x^n }{1-x^n }$$

my question is..if you know what S(x) is then..could you obtain the value of the a(n) ?..or simply working with the Dirichlet series version:

$$\sum_{n=0}^{\infty}a(n)n^{-s}= \zeta (s) \sum _{n=0}^{\infty}b(n)n^{-s}$$

Where $$S(x)= \sum_{n=0}^{\infty} \frac{ a(n) x^n }{1-x^n }=\sum_{n=0}^{\infty} \frac{ b(n) x^n }$$

2. Aug 25, 2006

saltydog

Notice this one still at zero so I'll say my two cents:

If you can represent S(x) as a taylor series, then the coefficients of the taylor series are related to the coefficients of the Lambert series; you can obtain the a(n) from those of the Taylor series through iteration. Check out Lambert series on Mathworld to see this and then try it for:

$$S(x)=\frac{x}{(1-x)^2}$$