Question

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

• A. Statement I is true; Statement II is true; Statement II is a correct explanation for Statement I
• B. Statement I is true; Statement II is true; Statement II is not a correct explanation for Statement I
• C. Statement I is true; Statement II is false
• D. Statement I is false; Statement II is true

Answer 1

Answer: (B)

I. Let $P(n):\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}>\sqrt{n}$, for all natural numbers $n\ge 2$.
$\therefore P(2):\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}=1.707>\sqrt{2}$
$\Rightarrow P(2)$ is true.
Let us assume that for $n=k,P(n)$ is true i.e.,
$P(k):\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{k}}>\sqrt{k}$ is true
Now for $n=k+1$, we have
$P(k+1):\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots +\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}$
$>\sqrt{k}+\frac{1}{\sqrt{k+1}}=\frac{\sqrt{k(k+1)}+1}{\sqrt{(k+1)}}$
$>\frac{k+1}{\sqrt{k+1}}[\because \sqrt{k(k+1)}>k,\forall k\ge 1]$
$=\sqrt{(k+1)}$
$\Rightarrow P(k+1):\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots +\frac{1}{\sqrt{k+1}}>\sqrt{k+1}$
$\therefore$ By principle of mathematical induction, Statement I is true, $\forall n\ge 2$.
II. Now, let $\alpha (n):\sqrt{n(n+1)}<n+1$, for all natural numbers $n\ge 2$.
$\therefore \alpha (2):\sqrt{2(2+1)}=\sqrt{6}<3$
Let us assume that
$\alpha (k):\sqrt{k(k+1)}<(k+1)$ is true
For $n=k+1$,
$\alpha (k+1):\sqrt{(k+1)(k+2)}<(k+2)[\because (k+1)<(k+2)]$
By principle of mathematical induction, Statement II is true, for all $n\ge 2$, but Statement II is not a correct explanation for Statement I.