Solution
= Solution
For the base case $n=2$, $1+1/\sqrt2>\sqrt2$, since multiplying by $\sqrt2$ reduces this to $\sqrt2+1>2$. Suppose the claimed inequality holds for some $n\geq2$. The identity
$$
\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}<\frac1{\sqrt{n+1}}
$$
then gives
$$
\sum_{k=1}^{n+1}\frac1{\sqrt k}
>\sqrt n+\frac1{\sqrt{n+1}}
>\sqrt{n+1}.
$$
Thus <mathematical induction> proves
$$
\boxed{\sqrt n<\sum_{k=1}^{n}\frac1{\sqrt k}\quad(n\geq2).}
$$