A physicist here so apologies in advance for lack of rigor, QFT, Path integrals etc ;)
Consider the function $f(w)$. I know independently (due to causality and other physical properties) that $f(w)$ should be analytic everywhere in the complex w plane, except for a branch cut that extends from $w=w_{min}$ to $w = \infty$ on the real axis. But $f(w)$ can, in principle, have a singularity at infinity.
For a particular physical set up, I can compute $f(w)$ as
$$f(w) = \sum_{n=1}^\infty a_n w^n \tag{1}$$
which converges for all $w<1$. I want to understand the properties of (1) as $w \rightarrow \infty$ in the complex $w$ plane. My goal is to make sure that $\frac{f(w)}{w}$ does not have a singularity at complex infinity in this particular set up.
Now $a_n$ is a complicated expression and it is almost impossible to do the above sum. But what if I do this — I know that $a_n$ monotonically decreases with $n$. So $a_1$ is the maximum of all $a_n$. I can then write,
$$f(w) < a_1 \sum_{n=1}^\infty w^n \tag{2}$$
Summing the geometric series and dividing by $w$ I get
$$ \frac{f(w)}{w} < \frac{a_1 }{1-w} \tag{3} $$
I can analytically continue beyond the pole at $w=1$ in the complex plane, and obtain that $f(w)/w$ vanishes at infinity and I am done.
The question is — Is the above correct?
The reason I am worried is because in the last step I analytically continued beyond $w=1$. And equation (2) does not make sense for $w>1$ sine the RHS is divergent.
I can imagine that this cannot be true for a generic function, since it would be quite magical if an inequality in a certain domain can be analytically continued over a bigger domain. However I was hoping that the analytic properties of $f(w)$ mentioned above could help in some way.
Thank you!