Is there an analytic function satisfying $\,\,f\big(\!\frac 1 n\!\big)=\frac 1 {\sqrt{n}}$ for all $n\in\mathbb N$? 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.
 A: We shall show that such analytic function does NOT exist. 
Assume the contrary, that such an analytic function $f :\Omega\to\mathbb C$ does exist, where $\Omega\subset \mathbb C$ is an open region , with $0\in\Omega$.
As $f$ is continuous at $z=0$, we have that
$$
f(0)=\lim_{n\to\infty}f\left(\frac{1}{n}\right)=\lim_{n\to\infty}\frac{1}{\sqrt{n}}=0.
$$
If we set $g(z)=\big(f(z)\big)^2$, then $g$ is also analytic in $\Omega$, and since
$$
g\Big(\frac{1}{n}\Big)=\frac{1}{\big(\sqrt{n}\big)^2}=\frac{1}{n},
$$ 
for all $n\in\mathbb N$, and $g(0)=0$, then, by virtue of the Identity Theorem $g$ has to be identical to the function $g(z)=z$, as the two functions agree in a set with a limit point in $\Omega$; Namely the agree on the set
$$
\left\{\frac{1}{n}:n\in\mathbb N\right\}\cup\big\{0\big\}.
$$
So $\big(f(z)\big)^2=z$. Differentiating we obtain
$$
2\,f(z)\,f'(z)=1,
$$
and setting $z=0$:
$$
0=2 \,f(0)\,f'(0)=1,
$$
as $\,f(0)=0$, which is a contradiction - we arrived to this contradiction having assumed that such $f$ existed.
A: 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 $f$.
