Question

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

• 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 $n \geq 2$.
$\therefore P(2): \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}=1.707>\sqrt{2}$
$\Rightarrow P(2)$ 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}}+\ldots+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}} $
$>\sqrt{k}+\frac{1}{\sqrt{k+1}}=\frac{\sqrt{k(k+1)}+1}{\sqrt{(k+1)}}$
$>\frac{k+1}{\sqrt{k+1}} [\because \sqrt{k(k+1)}>k, \forall k \geq 1]$
$=\sqrt{(k+1)} $
$ \Rightarrow P(k+1): \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{k+1}}>\sqrt{k+1}$
$\therefore$ By principle of mathematical induction, Statement I is true, $\forall n \geq 2$.
II. Now, let $\alpha(n): \sqrt{n(n+1)}< n+1$, for all natural numbers $n \geq 2$.
$\therefore \alpha(2): \sqrt{2(2+1)}=\sqrt{6}<3$
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 $n \geq 2$, but Statement II is not a correct explanation for Statement I.