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It may be a loss in translation, but I have been taught that a removable (effaçable in French) singularity for a power series lies necessarily within the interior of the convergence disc, yet I found this

Recall that $ℓ(t)=ℓ^{[0,1]}(t)$ has radius of convergence $r$ . Notice that $0<ℓ^J(t)≤ℓ(t)$ for all $t∈[0,r)$ so that the singularity of $ℓ^J(t)/ℓ(t)$ at $t=r$ is removable. The limit L$(J)=\lim_{t↗r}ℓ^J(t)ℓ(t)$ therefore exists and lies in $[0,1]$.
(Source, the $ℓ(t)$ and friends are formal power series)

I am a bit confused here, does removable only means that there is a continuous extension at $r$? But then, I can't see how it is inferred from the boundedness of $ℓ^J(t)/ℓ(t)$. Or is it something else?

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Maybe it's like the series for $1/(1-z)=\sum_{k\ge 0}z^k$. It has a radius of convergence $1$, yet the series can be defined at points $e^{i\theta}$, $\theta\in (0,2\pi)$. After that, we can find an analytical prolongation on $\mathbb C\setminus \{1\}$. –  TZakrevskiy Jul 25 '13 at 13:28
Power series do not have removable singularities. $f(z)=(1/z)\sin z$, for example, has a removable singularity at $z=0$, but the power series for $f(z)$ has no singularity there. You may find en.wikipedia.org/wiki/Removable_singularity informative. –  Gerry Myerson Jul 30 '13 at 11:17
@GerryMyerson Well, I know that, it is the point of this question. What could removable singularity mean in this context? –  Evpok Jul 30 '13 at 12:10
I suppose it means you have a fraction, and neither numerator nor denominator is defined at $t=r$, so there's a singularity --- but somehow the fraction tends to a limit as $t$ goes to $r$, so the singularity is removable. To give a silly example, $(\sum2x^n)/(\sum x^n)$ has a removable singularity at $x=1$. –  Gerry Myerson Jul 30 '13 at 13:00
@GerryMyerson Ah, is this called “removable singularity”, too? Then I suppose you have it right. There are different word for this and complex analysis singularities in French. –  Evpok Jul 30 '13 at 13:12

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