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Suppose $\langle z_n \rangle_{n\ge0}$ is a sequence of complex numbers such that $\sum_{n=0}^\infty z_n$ converges,Given that $f$ is an entire function such that $f(z_n)=n$, then

  1. $f\equiv 0$.

  2. $f$ is unbounded.

  3. No such $f$ exists.

  4. $f$ has no zeroes.

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Hint: convergence of the series implies convergence of its terms to zero. – Gerry Myerson Jun 5 '12 at 13:06
Yes.${}{}{}{}{}$ – Gerry Myerson Jun 5 '12 at 13:18
Look into Liouville's Theorem. You may find it helpful. – Michael Holland Jun 5 '12 at 13:19
@GerryMyerson $\lim_{n\to\infty}f(z_n)=n\Rightarrow f(0)=\infty$ so $f$ is unbounded. – Un Chien Andalou Jun 18 '13 at 8:14
It's unbounded, but it's unbounded on a sequence of inputs converging to zero, so it's not entire since it has no finite value at zero, so it doesn't exist. – Gerry Myerson Jun 18 '13 at 11:17
up vote 5 down vote accepted

As Gerry pointed out, $z_n \rightarrow 0$. Entire functions are continuous, so think about what we can say about $f(0)$.

Also, what you have written about $g(z)$ doesn't make any sense. You can't say $g(z) = f(z) - n$ has zeros at all of the $z_k$ because for the definition of $g(z)$ to make sense, $n$ must be fixed. So it would have a zero at $z_n$ but not at $z_k$ if $k \neq n$.

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$f(0)=0$, according to the question. – Un Chien Andalou Jun 5 '12 at 13:40
It doesn't say that anywhere in the question you posted. – Seth Jun 5 '12 at 13:40
oh! I see I am very sorry. – Un Chien Andalou Jun 5 '12 at 13:43
I must agree that I should take $g(z)=f(z)-z$ right?and then all $z_n$ will be zeroes of $g$. – Un Chien Andalou Jun 5 '12 at 13:45

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