TITLE: Is there an analytic function satisfying $\,\,f\big(\!\frac 1 n\!\big)=\frac 1 {\sqrt{n}}$ for all $n\in\mathbb N$?
QUESTION [11 upvotes]: Is there a function which is analytic in an open neighbourhood of $z=0$ and satisfies
$$
f\left(\!\dfrac 1 n\!\right)=\dfrac 1  {\sqrt{n}},
$$ 
for all natural numbers $n$? 
I guess this problem requires the Identity Theorem but I can't use it the natural way because 
the function $$f(z)=\sqrt z,$$ isn't analytic in $z=0$. I would like a hint.

REPLY [18 votes]: An alternative proof that only assumes the differentiability of $f$ at $0$:
If $f$ is differentiable it is continuous at $0$, so $f(0) = \lim f(\frac{1}{n}) = \lim \frac{1}{\sqrt{n}} = 0$. By definition $f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x}$, so in particular:
$$f'(0) = \lim \frac{f(1/n)}{1/n} = \lim \sqrt{n} = + \infty$$
Which contradicts the differentiability of $f$.