= Solution
Rationalize the numerator of the summand:
$$
\frac{\sqrt{n+1}-\sqrt n}{\sqrt n}
=\frac1{\sqrt n\,(\sqrt{n+1}+\sqrt n)}
=\frac1{n(\sqrt{1+1/n}+1)}.
$$
For every $n\ge1$ we have $\sqrt{1+1/n}\le\sqrt2$, and therefore
$$
\frac{\sqrt{n+1}-\sqrt n}{\sqrt n}
\ge\frac1{(\sqrt2+1)n}>0.
$$
The <harmonic series> diverges, so the <comparison test for series> makes the given positive-term series diverge as well:
$$
\boxed{\sum_{n=1}^\infty
\frac{\sqrt{n+1}-\sqrt n}{\sqrt n}=+\infty.}
$$
For completeness, divergence of the <harmonic series> follows by grouping the terms with $2^k<n\le2^{k+1}$: each such block contributes at least $2^k/2^{k+1}=1/2$. The same rationalization also shows that $n$ times the summand tends to $1/2$. \b[The summands tend to zero, but their harmonic-size tail prevents convergence.]
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