<s>[INST] Show that $\frac{1}{2}\le \sum\limits_{k=0}^n\frac{1}{n+k}\le1$ 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....
 [/INST] It holds when $k=1$
$$\frac { 1 }{ n+1 } +\frac { 1 }{ n+2 } +...+\frac { 1 }{ 2n } \ge \overset { n }{ \overbrace { \frac { 1 }{ 2n } +\frac { 1 }{ 2n } +...\frac { 1 }{ 2n }  }  } =n\frac { 1 }{ 2n } =\frac { 1 }{ 2 } \\ \\ \frac { 1 }{ n+1 } +\frac { 1 }{ n+2 } +...+\frac { 1 }{ 2n } \le \overset { n }{ \overbrace { \frac { 1 }{ n+1 } +\frac { 1 }{ n+1 } +...+\frac { 1 }{ n+1 }  }  } =\frac { n }{ n+1 } \le 1\\ $$
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