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## Main Question or Discussion Point

If we have the next Lambert series:

[tex] S(x)= \sum_{n=0}^{\infty} \frac{ a(n) x^n }{1-x^n } [/tex]

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:

[tex] \sum_{n=0}^{\infty}a(n)n^{-s}= \zeta (s) \sum _{n=0}^{\infty}b(n)n^{-s} [/tex]

Where [tex] S(x)= \sum_{n=0}^{\infty} \frac{ a(n) x^n }{1-x^n }=\sum_{n=0}^{\infty} \frac{ b(n) x^n } [/tex]

[tex] S(x)= \sum_{n=0}^{\infty} \frac{ a(n) x^n }{1-x^n } [/tex]

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:

[tex] \sum_{n=0}^{\infty}a(n)n^{-s}= \zeta (s) \sum _{n=0}^{\infty}b(n)n^{-s} [/tex]

Where [tex] S(x)= \sum_{n=0}^{\infty} \frac{ a(n) x^n }{1-x^n }=\sum_{n=0}^{\infty} \frac{ b(n) x^n } [/tex]