Consider strictly $n\geq 4$ (this won't affect the convergence of the series). Then $\sqrt{n}\leq \frac n2$ (why?); this means that you can say that $n-\sqrt{n}\geq \frac n2$. Likewise, you have that $n+\sqrt{n}\leq n+\frac n2 = \frac{3n}2$. Combine these and you get that $\frac{n-\sqrt{n}}{\left(n+\sqrt{n}\right)^2}\geq \dfrac{\frac n2}{\left(\frac{3n}2\right)^2} = \frac{2}{9n}$. Now use the divergence of the harmonic series.
The problem with your proposed solution is that you've misapplied the ratio test; you need to be able to say that $\left(\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\right)\lt 1$, not just that the limit is $\leq 1$ or even that $\left|\frac{a_{n+1}}{a_n}\right|\lt 1$ for all $n$. In fact, the harmonic series $\sum_{n\to\infty}\frac1n$ itself serves as a counterexample to your use of the limit test; with $a_n=\frac1n$, we have $\left|\frac{a_{n+1}}{a_n}\right|\lt 1$ for all $n$, $\lim_{n\to\infty} a_n=0$, and $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\leq 1$, but $\sum a_n$ diverges. In general, the harmonic series can serve as a useful general-purpose counterexample: whenever you're unsure of your application of a series convergence test, try asking yourself 'what happens if I apply my logic to the harmonic series?'.