Prove the statement by the Principle of Mathematical Induction: $\sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{n}}$ for all natural numbers $n \geq 2$.
(N/A) Let $P(n): \sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \ldots + \frac{1}{\sqrt{n}}$ for $n \geq 2$.
For $n=2$,$P(2): \sqrt{2} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} = 1 + \frac{1}{1.414} \approx 1 + 0.707 = 1.707$. Since $\sqrt{2} \approx 1.414$,$1.414 < 1.707$ is true.
Assume $P(k)$ is true for some $k \geq 2$: $\sqrt{k} < \sum_{i=1}^{k} \frac{1}{\sqrt{i}}$.
We need to show $P(k+1): \sqrt{k+1} < \sum_{i=1}^{k+1} \frac{1}{\sqrt{i}}$.
We know $\sqrt{k+1} - \sqrt{k} = \frac{(\sqrt{k+1} - \sqrt{k})(\sqrt{k+1} + \sqrt{k})}{\sqrt{k+1} + \sqrt{k}} = \frac{1}{\sqrt{k+1} + \sqrt{k}}$.
Since $\sqrt{k+1} + \sqrt{k} > 2\sqrt{k} > \sqrt{k+1}$,we have $\frac{1}{\sqrt{k+1} + \sqrt{k}} < \frac{1}{\sqrt{k+1}}$.
Thus,$\sqrt{k+1} < \sqrt{k} + \frac{1}{\sqrt{k+1}} < \sum_{i=1}^{k} \frac{1}{\sqrt{i}} + \frac{1}{\sqrt{k+1}} = \sum_{i=1}^{k+1} \frac{1}{\sqrt{i}}$.
Thus,$P(k+1)$ is true. By the Principle of Mathematical Induction,$P(n)$ is true for all $n \geq 2$.
For $n=2$,$P(2): \sqrt{2} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} = 1 + \frac{1}{1.414} \approx 1 + 0.707 = 1.707$. Since $\sqrt{2} \approx 1.414$,$1.414 < 1.707$ is true.
Assume $P(k)$ is true for some $k \geq 2$: $\sqrt{k} < \sum_{i=1}^{k} \frac{1}{\sqrt{i}}$.
We need to show $P(k+1): \sqrt{k+1} < \sum_{i=1}^{k+1} \frac{1}{\sqrt{i}}$.
We know $\sqrt{k+1} - \sqrt{k} = \frac{(\sqrt{k+1} - \sqrt{k})(\sqrt{k+1} + \sqrt{k})}{\sqrt{k+1} + \sqrt{k}} = \frac{1}{\sqrt{k+1} + \sqrt{k}}$.
Since $\sqrt{k+1} + \sqrt{k} > 2\sqrt{k} > \sqrt{k+1}$,we have $\frac{1}{\sqrt{k+1} + \sqrt{k}} < \frac{1}{\sqrt{k+1}}$.
Thus,$\sqrt{k+1} < \sqrt{k} + \frac{1}{\sqrt{k+1}} < \sum_{i=1}^{k} \frac{1}{\sqrt{i}} + \frac{1}{\sqrt{k+1}} = \sum_{i=1}^{k+1} \frac{1}{\sqrt{i}}$.
Thus,$P(k+1)$ is true. By the Principle of Mathematical Induction,$P(n)$ is true for all $n \geq 2$.
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