Prove that $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}\geq \frac{2}{3}$

Question

Prove that for $n \in \mathbb{N}$

(a) $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}\geq \frac{2}{3}$

And

(b) $\frac{1}{2} \leq \frac{1}{3n+1}+\frac{1}{3n+2}+.....\frac{1}{5n}+\frac{1}{5n+1}+....< \frac{2}{3}$

my attempt :

$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}\\ >\frac{1}{2n}+\frac{1}{2n}+\frac{1}{2n}+....+\frac{1}{2n}\\ =\frac{n+1}{2n}>\frac{1}{2}$

Where i am missing and how to processed second question

Answer 1

By C-S $$\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\geq\frac{(1+1+...+1)^2}{n+(n+1)+...+(2n)}=\frac{(n+1)^2}{\frac{(n+1)(2n+n)}{2}}=\frac{2(n+1)}{3n}>\frac{2}{3}$$

$$\frac{1}{3n+1}+\frac{1}{3n+2}+.....\frac{1}{5n}+\frac{1}{5n+1}\geq\frac{(1+1+...+1)^2}{(3n+1)+(3n+2)+...+(5n+1)}=$$ $$=\frac{(2n+1)^2}{\frac{(2n+1)(2(3n+1)+2n)}{2}}=\frac{2n+1}{4n+1}>\frac{1}{2}.$$ $$\frac{1}{3n+1}+\frac{1}{3n+2}+.....\frac{1}{5n}+\frac{1}{5n+1}<\int\limits_{3n}^{5n+1}\frac{1}{x}dx=\ln\frac{5n+1}{3n}<\frac{2}{3}$$ for all $n\geq2$.

For $n=1$ we see that $\frac{1}{4}+\frac{1}{5}+\frac{1}{6}<\frac{2}{3}$ is true.