# Is there a reason why the Maclaurin coefficients of the Riemann zeta function are asymptotically close to -1?

If I look at the numerical values of the Maclaurin series of the Riemann zeta function I see that they approach -1 extremely quickly. In fact, if I take $$\zeta(x)=\sum_{n=0}^\infty a_n x^n$$ then numerically I find $$\sum_{n=0}^\infty(a_n+1)^2\approx 0.25658169$$ which converges very quickly.

This seems to be stronger than just asking why $$-\frac{1}{2}+\frac{x}{x-1}$$ approximates $$\zeta(x)$$ so well.

Let $$f(z)$$ be any function with a simple pole at $$z=z_0$$ with residue $$r_0$$ and otherwise entire. Then $$g(z) = f(z) - r_0/(z-z_0)$$ is entire, and so in its Maclaurin expansion $$g(z) = \sum_{n=0}^\infty g_n z^n$$, the coefficients $$g_n$$ decay superexponentially (by the root test or by Cauchy's integral formula), $$b_n \ll_A A^{-n}$$ for every $$A>1$$. But then the Maclaurin expansion $$f(z) = \sum_{n=0}^\infty f_n z^n$$ is dominated by the behaviour of $$r_0/(z-z_0)$$: $$\sum_{n=0}^\infty f_n z^n = \frac{r_0}{z-z_0} + \sum_{n=0}^\infty g_n z^n = \frac{-r_0/z_0}{1-z/z_0} + \sum_{n=0}^\infty g_n z^n = \sum_{n=0}^\infty \frac{-r_0}{z_0} \biggl( \frac z{z_0} \biggr)^n + \sum_{n=0}^\infty g_n z^n,$$ and hence $$\displaystyle f_n = \frac{-r_0}{z_0^{n+1}} + O_A( A^{-n} ).$$