Proce that, for n > 1,

1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{n}} \gt \sqrt{n}.

Proof

The proof is by induction. For convenience, denote the left hand side as S_{n}.

Let n = 2. Then 1 + \frac{1}{\sqrt{2}} = 1 + \frac{\sqrt{2}}{2} > \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} because 1 > \frac{\sqrt{2}}{2}.

Assume that the inequality holds for n = k: S_{k} > \sqrt{k} and let n = k + 1. We have to prove that S_{k+1} > \sqrt{k+1}. But S_{k+1} = S_{k} + \frac{1}{\sqrt{k+1}}. So, by the inductive assumption,

S_{k+1} = S_{k} + \frac{1}{\sqrt{k+1}} > \sqrt{k} + \frac{1}{\sqrt{k+1}}.

We have an obvious inequality \sqrt{k(k+1)} > k = (k+1)-1. Divide this by \sqrt{k+1}:

\sqrt{k} > \sqrt{k+1}-\frac{1}{\sqrt{k+1}},

which leads exactly to the inequality needed for the inductive step:

S_{k+1} = S_{k} + \frac{1}{\sqrt{k+1}} > \sqrt{k} + \frac{1}{\sqrt{k+1}} > \sqrt{k+1}-\frac{1}{\sqrt{k+1}}+\frac{1}{\sqrt{k+1}} = \sqrt{k+1}.

Note The problem appears to be made to be resolved by the mathematical induction. But we were actually lucky. There is a series of apparently similar problems that could not be solved directly in this manner.

Amazingly, the strengthened inequalities readily yield to the inductive argument.