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 $k\geq1$,


by addition

$ 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 $n\geq2$ .

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

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

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

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.