# What is the relationship between the asymptotic series outside the radius of convergence and the analytic continuation?

In his post about computing the zeta function by smooth cutoff regularization of the series $\sum\frac{1}{n^2}$, Terry Tao shows how $-{1\over 12}$ is the finite part of the asymptotic series for the smooth cutoff regularization of the sum $\sum\frac{1}{n}$. In the last section of his post, he justifies why that value should match the analytic continuation of the function from its convergent domain. While all the steps of that argument make sense, I am somehow missing the larger picture. Can someone attempt another explanation?

Let's make my question more concrete. If $f(z)=\sum a_nz^n$ is absolutely convergent for $\lvert z\rvert < R$ with a pole at some $z_0$, $\lvert z_0\rvert = R$, and has analytic continuation $\tilde f(z)$ defined on $\mathbb{C}\setminus z_0$, is there some standard way to break $f$ into an asymptotic series, with a demarcated finite part and infinite part? Like $f(z)\sim\sum_{n=1}^N b_n\epsilon^{-n} +\sum_{n=0}^\infty c_n\epsilon^n$ as $\epsilon\to 0$, and with $\sum_{n=0}^\infty c_n\epsilon^n\to \tilde f(z)$. If we discard the divergent part, the finite part matches the analytic continuation. Why? This is what Tao does, and seems to also match the approach of some shady QFT maneuvers.