Hmm. If the above is true, then
\( \sum_{n=0}^{\infty} \frac{z^n}{n!} \zeta(-n) = \frac{1}{1 - e^z} \) (which is false)
but
\( \sum_{n=0}^{\infty} \frac{z^n}{n!} \zeta(-n)
= \frac{e^z}{1 - e^z} + \frac{1}{z}
= \frac{1 + e^z(z-1)}{z(1 - e^z)} \)
is the actual result.
So I must have made a slight mistake... I'll have to think on this.
However, a similar form gives:
\( \sum_{n=-1}^{\infty} \frac{z^n}{n!} \zeta(-n)
= \frac{e^z}{1 - e^z} \)
because \( \lim_{n \to -1} \frac{z^n\zeta(-n)}{n!} = -\frac{1}{z} \)
\( \sum_{n=0}^{\infty} \frac{z^n}{n!} \zeta(-n) = \frac{1}{1 - e^z} \) (which is false)
but
\( \sum_{n=0}^{\infty} \frac{z^n}{n!} \zeta(-n)
= \frac{e^z}{1 - e^z} + \frac{1}{z}
= \frac{1 + e^z(z-1)}{z(1 - e^z)} \)
is the actual result.
So I must have made a slight mistake... I'll have to think on this.
However, a similar form gives:
\( \sum_{n=-1}^{\infty} \frac{z^n}{n!} \zeta(-n)
= \frac{e^z}{1 - e^z} \)
because \( \lim_{n \to -1} \frac{z^n\zeta(-n)}{n!} = -\frac{1}{z} \)
