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Let $\varphi$ be the Euler totient function, and let us define the function $f(z)$ by the series $$ f(z) := \sum_{n=1}^{\infty} \varphi(n) z^n $$ Since $0\le \varphi(n)\le n$, I believe this gives a well-defined function in some region of the complex plane. Where could I find a discussion on the analytic behavior of this function as a function of the complex variable $z$?

I am in particular interested in the pole structure as we approach $z=1$ from below ($z\to 1^{-}$), say on the real axis. More specifically let us write $z=e^{-\beta}$. Then for $\beta$ positive I would expect the function $f(\beta)$ to converge. Then I am interested in the limit $\beta\to 0^+$ from above, and especially in the constant part $c_0$: $$ f(z=e^{-\beta})=\frac{c_n}{\beta^n} + \cdots +\frac{c_1}{\beta} +c_0 + c_{-1} \beta +\cdots . $$

PS: as suggested by the comment, a potential clue is that there is a known identity $$ g(s):=\sum_{n=1}^{\infty} \frac{\varphi(n)}{n^s} =\frac{\zeta(s-1)}{\zeta(s)} $$

PS^2: As is commented in the answers, it indeed seems to be the case that $f(z)$ has the unit circle as natural boundary. However we can still ask the question of what the value of $c_0$ is, at least in certain cases. As an example consider a different function, which indeed takes the form of a Lambert series: $$ h(z):=\sum_{n=1}^{\infty} \varphi(n) \frac{z^n}{1-z^n} $$ This has unit circle as natural boundary. But then there is a nice result that this can be written as $$ h(z)=\frac{z}{(1-z)^2} $$ and hence $$ h(z=e^{-\beta})=\frac{1}{\beta^2}-\frac{1}{12}+\cdots $$ so we learn that $c_0=-\frac{1}{12}$ in this case. What I mean is that I would like to extract a similar number $c_0$ for the function $f(z)$. It indeed is not clear if this question has a well-defined answer for $f(z)$.

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  • $\begingroup$ This is the inverse Mellin transform of $\Gamma(s)\zeta(s-1)/\zeta(s)$. $\endgroup$ – M.G. Apr 27 '15 at 22:07
  • $\begingroup$ Defined for $|z|<1$ certainly, but $|z|=1$ might be a natural boundary, in which case there's no pole at $z=1$ and thus no Laurent expansion about $z=1$. (There could still be an asymptotic expansion as $z$ approaches $1$ from below on the real axis.) $\endgroup$ – Noam D. Elkies Apr 27 '15 at 23:25
  • $\begingroup$ Thank you, in fact I only need to consider $z=e^{-\beta}$, with $\beta$ real and positive, and the limit is $\beta\to 0$. I modified the text accordingly. $\endgroup$ – user16070 Apr 27 '15 at 23:56
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Note that $f(z)=\sum_{n=1}^\infty \phi(n) z^n$ can be written as a Lambert series, $$f(z) = \sum a_n\frac{z^n}{1-z^n},$$ where $a_n = \sum_{d|n}\phi(d) \mu(n/d).$ It is clear that the Lambert series converges inside the unit disk, and $f(z)$ has a natural boundary on the unit circle (since it will have a pole at every root of unity).

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  • $\begingroup$ Thank you so much! This makes it clear that we have a natural boundary on the unit circle. However I still think my original question remains. To formulate the questions sharper, I updated the question above (please see the paragraph: PS^2). $\endgroup$ – user16070 Apr 28 '15 at 5:05
  • $\begingroup$ Forgive me if this a silly question, but how does one directly show that $f$ has a singular point at every root of unity? $\endgroup$ – Matt Rosenzweig Jul 4 '15 at 3:15
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F. Carlson proved in 1921 that a power series with integer coefficients and radius of convergence 1 is either a rational function or has the unit circle as a natural boundary. Since $f(z)$ is clearly not rational, it has the unit circle as a natural boundary. For some references, see the solution to Exercise 4.46(c) of Enumerative Combinatorics, vol. 1, 2nd ed.

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  • $\begingroup$ Thank you for the comment. I will go to the library and get the book tomorrow. I tried to better formulate the question this time, please see the updated question above (please see the paragraph: PS^2). $\endgroup$ – user16070 Apr 28 '15 at 5:08

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