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\begin{equation} f(z) = \sum_{n=1}^\infty \frac{z^n}{n^2} \end{equation}

Has a radius of convergence R = 1, actually it converges for $|z| \leq 1$ but as you know, the radius of convergence is the distance to the nearest singularity, this singularity prevents the radius to be even bigger

But what is this singularity? I can't see it

GodrickTheGrafted
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    The series diverges for $z=1+\epsilon$ for any $\epsilon > 0$. – angryavian Apr 01 '22 at 02:09
  • $f'(r) \to \infty, r \to 1$ so $1$ is a singular point of $f'$ hence of $f$ – Conrad Apr 01 '22 at 02:15
  • Thanks for your comments, Conrad, how does $f'(r) \to \infty, r \to 1$ imply a singular point on $f$? I mean, we can have singular points in derivatives and defined functions right? for example $x^{1/3}$, the derivative at $x = 0$ is divergent but the function is defined – GodrickTheGrafted Apr 01 '22 at 02:44
  • @IsaacGuillenCastellanos The function is not holomorphic at $1$ since it is not complex differentiable in a neighborhood of $1$ (since the derivative blows up there). When one says "distance to the nearest singularity", it's essentially the distance to the nearest point where your function can't be holomorphically extended starting from some neighborhood of your starting point. Alternatively, angryavian's comment is a bit more direct. – Brevan Ellefsen Apr 01 '22 at 05:47
  • I see... it makes sense, according to wikipedie: "that is complex differentiable in a neighbourhood of each point in a domain" could you say that it is a removable singularity? thanks! – GodrickTheGrafted Apr 01 '22 at 23:38

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