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 $f(z)$.



Residue for the generating function of the Euler totient function
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).


