Does $\sum\limits_{n=1}^\infty\frac{1}{\sqrt{n}+\sqrt{n+1}}$ converge?

For $n\geq 1$, we have $\sqrt n+\sqrt{n+1}\leq 2\sqrt{n+1}\leq 2(n+1)\leq 4n$ hence $$\frac 1{\sqrt n+\sqrt{n+1}}\geq \frac 1{4n}\geq 0$$ and we can conclude using the fact that the harmonic series $\sum_{k=1}^{+\infty}\frac 1k$ is divergent.

It is not hard to see that $$\sum_{n=1}^\infty\frac{1}{\sqrt{n+1}+\sqrt{n}}=\sum_{n=1}^\infty(\sqrt{n+1}-\sqrt{n})$$

As you know this series is divergent.

You also have all of your experience with limits and approximation available. The key observation that makes things 'obvious' is that $\sqrt{n+1} \approx \sqrt{n}$, and so

$$ \frac{1}{\sqrt{n} + \sqrt{n+1}} \approx \frac{1}{2\sqrt{n}} $$

and so you can apply your knowledge about the convergence of sums of the form $\sum 1/n^s$. For example, since $s = 1/2$, this should diverge faster than the harmonic series - a lot faster really - and so you should have no trouble comparing the original sum to the harmonic series (e.g. as in Davide's answer).