show that $\frac{1}{3n+1}+\frac{1}{3n+2}+...+\frac{1}{5n}+\frac{1}{5n+1} < \frac{2}{3}$ , $\forall n \mathbb \in{N}$

Question

Show that $$\frac{1}{3n+1}+\frac{1}{3n+2}+...+\frac{1}{5n}+\frac{1}{5n+1} < \frac{2}{3}$$ for all $n \in \Bbb{N}$

I tried with induction method but I can not find any results.

Answer 1

Hint:

If $f(m)=\sum_{r=3m+1}^{5m+1}\dfrac1r,$

It is sufficient to show $f(n)\ge f(n+1)$ and $f(1)<\dfrac23$

Answer 2

$$\frac{1}{3n+1}+\frac{1}{3n+2}+\cdots+\frac{1}{5n+1} <\frac{1}{3n+1}+\frac{1}{3n+2}+\frac{1}{3n+2}+\cdots+\frac{1}{3n+2}$$ $$=\frac{1}{3n+1}+\frac{2n}{3n+2}<\frac{2}{3}$$