Statement-1 : For every natural number $n \geq 2$, $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\fr

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Statement-1 : For every natural number $n \geq 2$, $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}}>\sqrt{n}$ Statement-2 : For every natural number $n \geq 2$, $\sqrt{n(n+1)}<n+1$.

Professor PC Thomas Classes · Answer

(d) : Statement-1 : Let\ $P(n): \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}}>\sqrt{n}$\ Step 1: For $n=2, P(2): \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}>\sqrt{2}$ is true\ Step 2 : Assume $P(n)$ is true for $n=k$, i.e. $$\begin{equation*}\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{k}}>\sqrt{k} ag{i} \end{equation*}$$ Step 3 : For $n=k+1$, we have to show that $$\begin{equation*}\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}} ag{ii} \end{equation*}$$ $>\sqrt{k+1}$\ By assumption step, we get\ $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{k}}>\sqrt{k}$\ Adding $\frac{1}{\sqrt{k+1}}$ on both sides, we get $$\begin{equation*}\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}} \sqrt{k}+\frac{1}{\sqrt{k+1}} ag{iii} \end{equation*}$$ Statement-2 : For $n=k, \sqrt{k(k+1)}\frac{\sqrt{k}}{\sqrt{k+1}} \Rightarrow \sqrt{k}>\frac{k}{\sqrt{k+1}}$ $$\begin{equation*}\Rightarrow \sqrt{k}+\frac{1}{\sqrt{k+1}}>\sqrt{k+1} ag{iv} \end{equation*}$$ From (iii) \& (iv)\ $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}} \sqrt{k+1}$\ Hence (ii) is true for $n=k+1$. Hence $P(n)$ is true for $n \geq 2$ So, Statement-1 and Statement-2 are correct but Statement-2 is not a correct explanation of Statement-1.\