# Prove the inequality for n $\in \mathbb{N}$

Prove the inequality for $n \in\mathbb{N}, n>1$

$\sqrt{n} < 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} +...+ \frac{1}{\sqrt{n}} < 2\sqrt{n}$

I know how to prove that it is bigger than $\sqrt{n}$ - just show it is bigger than $\frac{1}{\sqrt{n}}\cdot n$, but the other thing is much more complicated.

Proof of RIGHT inequality

$t \mapsto \frac{1}{\sqrt{t}}$ is decreasing, so for each $k\geq1$,

$\int_k^{k+1}\frac{dt}{\sqrt{t}}\leq\frac{1}{\sqrt{k}}\leq\int_{k-1}^{k}\frac{dt}{\sqrt{t}}$

$2(\sqrt{n+1}-1)\leq\sum_{k=1}^n\frac{1}{\sqrt{k}} \leq 2\sqrt{n}.$

let now prove the left inequality.

if $n=1$, it is satisfied.

assume $n\geq2$ .

$n\geq2 \implies 4\leq 3\sqrt{n}$

$\implies n+4+4\sqrt{n}\leq 4n+4$

$\implies (\sqrt{n}+2)^2\leq 4(n+1)$

$\implies \sqrt{n}+2\leq2\sqrt{n+1}$

$\implies \sqrt{n}\leq \sum_{k=1}^{n}\frac{1}{\sqrt{k}}$ .