# Show that 1+1/sqrt2+cdots+ 1/sqrtn >=sqrt2(n-1), for n >1?

Jul 15, 2018

Below

#### Explanation:

To show that the inequality is true, you use mathematical induction

$1 + \frac{1}{\sqrt{2}} + \ldots + \frac{1}{\sqrt{n}} \ge \sqrt{2} \left(n - 1\right)$ for $n > 1$

Step 1: Prove true for $n = 2$
LHS=$1 + \frac{1}{\sqrt{2}}$
RHS=$\sqrt{2} \left(2 - 1\right) = \sqrt{2}$
Since $1 + \frac{1}{\sqrt{2}} > \sqrt{2}$, then $L H S > R H S$. Therefore, it is true for $n = 2$

Step 2: Assume true for $n = k$ where k is an integer and $k > 1$

$1 + \frac{1}{\sqrt{2}} + \ldots + \frac{1}{\sqrt{k}} \ge \sqrt{2} \left(k - 1\right)$ --- (1)

Step 3: When $n = k + 1$,
RTP: $1 + \frac{1}{\sqrt{2}} + \ldots + \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k + 1}} \ge \sqrt{2} \left(k + 1 - 1\right)$
ie $0 \ge \sqrt{2} - \left(1 + \frac{1}{\sqrt{2}} + \ldots + \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k + 1}}\right)$

RHS
=$\sqrt{2} - \left(1 + \frac{1}{\sqrt{2}} + \ldots + \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k + 1}}\right)$
$\ge \sqrt{2} - \left(\sqrt{2} \left(k - 1\right) + \frac{1}{\sqrt{k + 1}}\right)$ from (1) by assumption
=$\sqrt{2} - \sqrt{2} \left(k\right) + \sqrt{2} - \frac{1}{\sqrt{k + 1}}$
=$2 \sqrt{2} - \sqrt{2} \left(k\right) - \frac{1}{\sqrt{k + 1}}$

Since $k > 1$, then $- \frac{1}{\sqrt{k + 1}} < 0$ and since $k \sqrt{2} \ge 2 \sqrt{2} > 0$, then $2 \sqrt{2} - k \sqrt{2} < 0$ so $2 \sqrt{2} - \sqrt{2} \left(k\right) - \frac{1}{\sqrt{k + 1}} = < 0$
=LHS

Step 4: By proof of mathematical induction, this inequality is true for all integers $n$ greater than $1$

Jul 15, 2018

The inequality as stated is false.

Eg, for $n = 3$:

${\underbrace{1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}}}}_{\approx 2.3} \cancel{\ge} \underbrace{\sqrt{2} \left(3 - 1\right)} {\setminus}_{\approx 2.8}$