A ratio test that fails, and how to conclude

The ratio test is inconclusive.
$$\frac{u_{n+1}}{u_n}=\frac{(2n+2)(2n+1)}{4(n+1)^2}=\frac{2n+1}{2n+2}\longrightarrow1.$$

A lower bound. We prove $u_n\ge\frac1{2\sqrt n}$ by induction. For $n=1$: $u_1=\frac12$. Assume it for $n$. Then
$$u_{n+1}=\frac{2n+1}{2n+2}u_n\ge\frac{2n+1}{2n+2}\cdot\frac1{2\sqrt n}\ge\frac1{2\sqrt{n+1}},$$
where the last inequality is equivalent (squaring) to $(2n+1)^2\ge4n(n+1)$, i.e. $1\ge0$.

Conclusion. $\sum\frac1{2\sqrt n}$ diverges ($p=\frac12$), so $\sum u_n$ diverges by comparison.

With Stirling. $u_n\sim\frac1{\sqrt{\pi n}}$, and the series diverges for the same reason.

Remark. Note that $u_n\to0$ yet $\sum u_n=\infty$. The series of $u_n$ is the Taylor series of $\frac1{\sqrt{1-x}}$ at $x=1$.