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Let $f$ be an entire function. Assume that $\mid f(1/n)\mid\le e^{-n}$ for all $n\in \mathbb{N}$. Show that $f$ vanishes identically.

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Are you sure it's not $|f(n^{-1}|\leq e^{-n}$? Otherwise take $f(z)=e^{-1}$. Hint: you can show that $f(0)=0$, then by induction that $f^{(n)}(0)=0$. – Davide Giraudo Oct 8 '12 at 12:35
Oh, yes, sorry for mistake, I have corrected it. Thanks for hint, I will try it. – StudentMath Oct 8 '12 at 12:49

Taking the limit as $n\to\infty$, you will find that $f(0)=0$. If $f$is not identically zero, you can write $$f(z)=\sum_{k=n}^\infty a_kz^k$$ with $n\ge1$ and $a_n\ne0$ (I trust you know why). Now consider $$\lim_{z\to0}\frac{f(z)}{z^n}$$ in general, and compare with the special case $z=1/n$ as $n\to\infty$.

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Thanks, got it $ lim$ goes to zero, from other side it should be equal to $a_{n}$ which is equal to $f^{n}(0)/n!$ – StudentMath Oct 8 '12 at 14:47

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