Statement - 1: For every natural number n ≥ 2, 1 1 + 1 2 + + 1 n >...

$$\begin{aligned}& P(n)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}}\\ & P(2)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}>\sqrt{2} \end{aligned}$$$Let us assume that$P(k)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{k}} \gt \sqrt{k}$is true$\therefore P(k+1)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}} \gt \sqrt{k+1}$has to be true. L.H.S.$ \gt \sqrt{\mathrm{k}}+\frac{1}{\sqrt{\mathrm{k}+1}}=\frac{\sqrt{\mathrm{k}(\mathrm{k}+1)}+1}{\sqrt{\mathrm{k}+1}}$Since$\sqrt{k(k+1)} \gt k \quad(\forall k \geq 0)$\therefore$\frac{\sqrt{\mathrm{k}(\mathrm{k}+1)}+1}{\sqrt{\mathrm{k}+1}}$>$\frac{\mathrm{k}+1}{\sqrt{\mathrm{k}+1}}=\sqrt{\mathrm{k}+1}Let $P(n)=\sqrt{n(n+1)} \lt n+1$ Statement $-1$ is correct. P(2)=$\sqrt{2 \times 3}$< 3 If $P(k)=\sqrt{k(k+1)} \lt (k+1)$ is true Now $P(k+1)=\sqrt{(k+1)(k+2)} \lt k+2$ has to be true Since $(k+1) \lt k+2$ \therefore $\sqrt{(\mathrm{k}+1)(\mathrm{k}+2)}$< ($\mathrm{k}+$2) Hence Statement $-2$ is not a correct explanation of Statement $-1$.