If $f$ is analytic and $|f(1/n)| < e^{-n}$ for all $n > 0$, why is $f(z) = 0$ for all $z$? Let $f$ be analytic on a connected open set $W$. We suppose that $0 \in W$ and $|f(1/n)| < e^{-n}$ for all $n > 0$. How shall I prove that $f(z) = 0$ on $W$?
I'm unsure how to approach this problem.
 A: Suppose that $f$ is not identically zero.  Since $f(1/n)<e^{-n}$, we must have $\lim_{z\to 0}f(z)=0$, and so the Laurent series of $f$ at $0$ is actually a Taylor series.  We can factor out the lowest power of $z$ appearing with a non-zero coefficient to write $f(z)=cz^kg(z)$ where $g(z)$ is an analytic function with $g(0)=1$ and $c$ is a nonzero constant.  By continuity, we can find $\epsilon>0$ where if $|z|<\epsilon, $1/2<|g(z)|$.
For any $n>1/\epsilon$, we have that $|f(1/n)|=|cn^{-k}g(1/n)|>|n^{-k}| |c/2|$, while simultanously $|f(1/n)|<e^{-n}$.  Therefore, we have to have $|n^{-k}| |c/2|<e^{-n}$ for all sufficiently large values of $n$.  This is impossible.
A: You simply prove $\dfrac{d^k f}{d x^k}\Big|_{x=0} = 0$.
If you know some numerical finite difference scheme, you can prove that
there exists real(rational) numbers $A_i$ that,
$\dfrac{d^k f}{d x^k}\Big|_{x=0} = \lim_{n\to\infty} n^k\sum A_i f(\dfrac{i}{n})$,
while exponetial growth is always faster than polynomial growth, the above limit will go to 0.

Btw. To solve for the coefficients $A_i$, you simply use Taylor expansion for 
$f(\dfrac{i}{n}) = \sum_{m=0}^{k-1} \dfrac{1}{m!} f^{(m)}(0)\cdot (\dfrac{i}{n})^m + \dfrac{1}{k!}f^{(m)}(0) (\dfrac{i}{n})^k + O(\dfrac{i}{n})^{k+1}$ 
thus we can solve $A_i$ from a system,
$\sum A_i (\dfrac{i}{n})^m = 0$ for $i=0, 1, 2, \cdots k - 1$
the system is a Vandermon, thus non-singular, it is solvable.
$\sum A_i (\dfrac{i}{n})^{k} = 1$ 
