1. The problem is to test the convergence of the series $$\sum_{n=1}^\infty \frac{\sqrt{n+1} - \sqrt{n}}{n^p}$$ for some real number $p$. 2. Recall the formula for the difference of square roots: $$\sqrt{n+1} - \sqrt{n} = \frac{(n+1) - n}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}}.$$ This helps simplify the terms of the series. 3. Substitute this into the series term: $$\frac{\sqrt{n+1} - \sqrt{n}}{n^p} = \frac{1}{(\sqrt{n+1} + \sqrt{n}) n^p}.$$ 4. For large $n$, $\sqrt{n+1} + \sqrt{n} \approx 2\sqrt{n}$, so the term behaves like $$\frac{1}{2 \sqrt{n} n^p} = \frac{1}{2 n^{p + \frac{1}{2}}}.$$ 5. The series behaves like $$\sum_{n=1}^\infty \frac{1}{n^{p + \frac{1}{2}}}$$ for large $n$. 6. We know the $p$-series $$\sum \frac{1}{n^q}$$ converges if and only if $q > 1$. 7. Therefore, the original series converges if and only if $$p + \frac{1}{2} > 1 \implies p > \frac{1}{2}.$$ 8. In conclusion, the series $$\sum_{n=1}^\infty \frac{\sqrt{n+1} - \sqrt{n}}{n^p}$$ converges if and only if $$p > \frac{1}{2}.$$