Inspired by another question, I got curious how we can calculate the radius of convergence for the formal power series:
$$f(x) = x\sqrt1+x^2\sqrt2+x^3\sqrt3+\cdots+x^n\sqrt{n}+\cdots$$
It is easy to see that it can not be larger than $1$, but how to find it?
If $\left|\frac{a_{n+1}}{a_n}\right|$ has a limit $\ell\in [0, \infty]$, the radius of convergence of the series $\sum_{n\ge 0} a_n z^n$ is $R = \frac{1}{\ell}$. In your case $R=1$ because $$\lim_{n\to \infty}\frac{\sqrt{n+1}}{\sqrt{n}}\to 1$$
Another solution here is to apply the Cauchy-Hadamard theorem.
Hadamard's formula: $$\frac1 R=\limsup_{n\to\infty}a_n^{\tfrac1n}=\limsup_{n\to\infty} n^{\tfrac1{2n}}=\limsup_{n\to\infty} \mathrm e^{\tfrac{\log n}{2n}}=\mathrm e^0.$$
