Question

$\text{Statement I}$ For every natural number $n\ge 2$
$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}>\sqrt{n}$
$\text{Statement II}$ For every natural number $n\ge 2$
$\sqrt{n(n+1)}<n+1$

• A. Statement $\text{I}$ is true, Statement $\text{II}$ is true; and Statement $\text{II}$ is correct explanation for Statement $\text{I}.$
• B. Statement $\text{I}$ is true, Statement $\text{II}$ is true; and Statement $\text{II}$ is not correct explanation for Statement $\text{I}.$
• C. Statement $\text{I}$ is true, Statement $\text{II}$ is false.
• D. Statement $\text{I}$ is false, Statement $\text{II}$ is true.

Answer 1

The correct option is A Statement $\text{I}$ is true, Statement $\text{II}$ is true; and Statement $\text{II}$ is correct explanation for Statement $\text{I}.$

For every natural number $n,$ we have
$n(n+1)=n^2+n<n^2+n+n+1$
$\Rightarrow n(n+1)<(n+1{)}^2$
$\Rightarrow \sqrt{n(n+1)}<(n+1)\forall n\ge 2$
$\therefore$ Statement $\text{II}$ is true.
Also, from above, we have
$\sqrt{n}<\sqrt{n+1}$
$\Rightarrow \frac{1}{\sqrt{n}}>\frac{1}{\sqrt{n+1}}\forall n\ge 2$
$\Rightarrow \frac{1}{\sqrt{1}}>\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{3}}>\cdots >\frac{1}{\sqrt{n-1}}>\frac{1}{\sqrt{n}}\forall n\ge 2$
$\Rightarrow \frac{1}{\sqrt{1}}>\frac{1}{\sqrt{n}}$
$\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{n}}$
$\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{n}}$
$\dots \dots \dots$
$\dots \dots \dots$
$\frac{1}{\sqrt{n}}=\frac{1}{\sqrt{n}},\forall n\ge 2$
Adding all, we get
$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots +\frac{1}{\sqrt{n}}>\frac{n}{\sqrt{n}}=\sqrt{n}$
$\therefore$ Statement $\text{I}$ is true.