Asymptotic Expression for $ f(z) = z+ z^\frac{1}{2}+ z^\frac{1}{3}+ z^\frac{1}{4} +\dots + z^\frac{1}{N}$ with complex $z$?

Question

Question (corrected)

I managed to prove:

$$ f(z) \sim \left\{ \begin{array}{ll} - \ln |z| \int_0^{\frac{-N}{\ln|z|}} e^{-\frac{1}{|y|}} dy & |z|<< 1 \\ ? & |z| \approx 1 \\ ?? & |z|\gg 1 \\ \end{array} \right. $$

Where, $$ f(z) = z+ z^\frac{1}{2}+ z^\frac{1}{3}+ z^\frac{1}{4} +\dots + z^\frac{1}{N}$$

However, I do not have a asymptotic expression as $|z| \to 1$ or $|z| \to \infty$. Can someone provide that expressions as well?

Background

Consider the complex function:

$$ f(z) = z+ z^\frac{1}{2}+ z^\frac{1}{3}+ z^\frac{1}{4} +\dots + z^\frac{1}{N}$$

Let, us write $z= r e^{i \theta}$ where $1>> r > 0$.

$$ f(r e^{i \theta}) = r e^{i \theta}+ r^{1/2}e^{i \theta/2} + \dots +r^{1/n}e^{i \theta/n}$$

Consider, the function with $f(x) = e^\frac{-1}{|x|}$ and the integral:

$$ \int_{0}^{N \epsilon} f(y) dy = \lim_{\epsilon \to 0,N \to \infty} \Big(f(\epsilon) +f(2 \epsilon) + \dots + f(N\epsilon) \Big)\epsilon$$

with $N \epsilon = b$. We modify our considerations and use$^\ast$:

$$ \lim_{\epsilon \to 0,N \to \infty} \sum_{r=1}^{N} a_r f(r\epsilon) \epsilon = \lim_{s\to 1} \frac{1}{\zeta(s)} \times \sum_{r=1}^\infty \frac{a_r}{r^s} \int_0^{N\epsilon} f(y) dy$$

Choosing $a_r= e^{i\theta /r}$ and replacing $\epsilon = \frac{-1}{\ln \delta}$

$$ \lim_{\delta \to 0,N \to \infty} \Big(f(\frac{-1}{\ln \delta})e^{i \theta} + f(\frac{-2}{\ln \delta})e^{i\theta /2} + \dots + f(\frac{-N}{\ln\delta})e^{i \theta/N} \Big) \frac{-1}{{\ln\delta}} = \underbrace{\lim_{s\to 1} \frac{1}{\zeta(s)} \times\sum_{r=1}^\infty \frac{e^{i\theta /r}}{r^s}}_{=1} \int_{0}^{- \frac{N}{\ln \delta}} e^\frac{-1}{|y|} dy$$

Substituting with $f$, using asymptotics and solving the limit using this:

$$ \delta e^{i \theta} + \delta^{1/2} e^{i \theta/2} + \dots + \delta^{1/N} e^{i \theta/N} \sim - \ln \delta \int_0^{\frac{-N}{\ln \delta}} e^{-\frac{1}{|y|}} dy $$

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^{$^\ast$We split into real and imaginary parts to apply the formula.}

Answer 1

As $z$ approaches $1$, the value of $f$ clearly approaches $N$. To see the difference term, we observe that as $z\rightarrow 1$, we have $z^p = (1 + (z-1))^p = 1 + p(z-1) + O(|z-1|^2)$ by the binomial series. Hence $$ f(z) - N = \sum_{n=1}^N (z^{1/N} - 1) = \sum_{n=1}^N \frac1n(z-1) + O(|z-1|^2) = (H_N)(z-1) + O(|z-1|^2) $$ where $H_N$ is the $N$th harmonic number. One can use the binomial series similarly to get higher order terms if you want. Since the function is differentiable, for any point you will have more generally as $z\rightarrow c$, $f(z)-f(c) = f'(c) (z-c) + O(|z-c|^2)$.

For historical reasons this is the asymptotics in terms of $N$: We have $$ z^{1/N} = \sum_{k=0}^\infty \frac{(\log z)^k}{k!} \frac1{N^k} = 1 + \frac{\log z}{N} +O(1/N^2) $$ hence we have $$ \sum_{n=1}^N z^{1/n} = N + (\log z)\sum_{n=1}^N \frac1n +O(1) = N + (\log z)(\log N) + O(1) $$