Proce that, for n > 1,

**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,

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

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

**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.

- \frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \ldots \cdot\frac{99}{100} < \frac{1}{10}
- 1 + \frac{1}{4} + \frac{1}{9} + \ldots < 2
- (1 + 1^{3})(1 + 2^{-3})(1 + 3^{-3}) + \ldots + (1 + n^{-3}) < 3.

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