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Is there an example of an analytic function in the unit disc whose zeros are only the points $z_n=1-1/n$?

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If there is such a function it should be $\prod_{n=1}^\infty (z-z_n)$. – azarel Apr 16 '12 at 20:06
@azarel: that doesn't converge. – GEdgar Apr 16 '12 at 20:14
@GEdgar Thanks for pointing that out. – azarel Apr 16 '12 at 20:25

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Another one: $$ f(z) = \frac{1}{\Gamma\left(\frac{1}{z-1}\right)} $$ This is analytic in the plane, except one point $z=1$, and has zeros exactly $1-1/n$, $n=1,2,3\dots$. Unlike Henning's, which also has zeros ${} \gt 1$.

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Thanks for the examples given, but how did you come to it? Maybe Henning's could have come to mind, but how did you think about this in general? – balestrav Apr 16 '12 at 21:34
I think it's just a matter of already knowing $1/\Gamma(z)$ as a standard example of a holomorphic function whose zeroes form a singly infinite arithmetic sequence. – Henning Makholm Apr 16 '12 at 22:03
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How about $f(z) = \sin(\frac{\pi}{1-z})$?

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For the general question of functions with prescribed zeros, consider Weierstraß' factorization theorem.

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Weierstrass factorization theorem is usually stated for entire functions, though. – mrf Apr 16 '12 at 20:15
True - I refered to this sentence from the Wikipedia page: "The theorem generalizes to the following: sequences in open subsets (and hence regions) of the Riemann sphere have associated functions that are holomorphic in those subsets and have zeroes at the points of the sequence." – Dirk Apr 16 '12 at 20:18
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Just use a fractional linear transformation to go from a question about a sequence of prescribed zeros with a finite limit $\alpha$ to a sequence of prescribed zeros with a limit of $\infty$: get an entire function, and transform it back (obtaining a function analytic on ${\mathbb C} \backslash \{ \alpha \}$). – Robert Israel Apr 16 '12 at 20:47

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