TITLE: Show that $\frac{1}{2}\le \sum\limits_{k=0}^n\frac{1}{n+k}\le1$
QUESTION [6 upvotes]: for $n\in \mathbb N^+$
$$\frac{1}{2n}\le \frac{\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{n+n}}{n}\le\frac{1}{n}$$
I tried math induction and I tried take integral but I want to solve this with most elementary methods  please give me hint or just show that. Thanks....

REPLY [4 votes]: I don't know if it is simple for you, but is a way. Using Abel's summation we get $$S=\sum_{k=0}^{n}\frac{1}{n+k}=\frac{1}{2}+\frac{1}{2n}+\int_{0}^{n}\frac{\left\lfloor t\right\rfloor +1}{\left(n+t\right)^{2}}dt
 $$ $$=\frac{1}{2}+\frac{1}{n}+\int_{0}^{n}\frac{\left\lfloor t\right\rfloor }{\left(n+t\right)^{2}}dt
 $$ where $\left\lfloor t\right\rfloor 
 $ is the floor function and since $t-1\leq\left\lfloor t\right\rfloor \leq t
 $ we get $$\frac{1}{2n}+\log\left(2\right)\leq S\leq\frac{1}{n}+\log\left(2\right)
 $$ hence the claim if $n\geq3$.