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Let $f$ be analytic in the disk $D(0,2)$ except for a pole of order $1$ at $z=1$, and let $$f(z)=\sum_{k=0}^\infty a_k z^k$$ be the series expansion for $f$ in the disk $D(0,1)$. Prove that $$\lim_{k \to \infty}\frac{a_{k+1}}{a_k}=1.$$

Intuitively, I want to say that the information given yields a Laurent series representation on $D(0,2)-\{1\}$:$$f(z)=\sum_{k=-1}^\infty b_k (z-1)^k$$ Expanding instead around $z=0$, we have an issue of a non-removable singularity at $z=1$ with analyticity everywhere else. Thus, the distance from $z=0$ to $z=1$ (the nearest singularity from the point of analyticity) is 1, so the result must follow by the Ratio Test as applied to the radius of convergence.

My solution has a feel of lack of rigor, however, and I don't know how to improve on its statement (or whether my approach is even valid).

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marked as duplicate by mrf, Claude Leibovici, achille hui, Nicholas R. Peterson, TMM Jan 7 at 9:35

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1 Answer 1

Hint: if you subtract a suitable multiple of $1/(1-z)$, what do you get?

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