Is there an analytic function satisfying $\,\,f\big(\!\frac 1 n\!\big)=\frac 1 {\sqrt{n}}$ for all $n\in\mathbb N$?

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.

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 $f$.