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....
 A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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The sum must be up to $\ds{\pars{n - 1}}$ with $\pars{n \geq 1}$:

\begin{align}
\begin{array}{rcccl}
\ds{\sum_{k = 0}^{n - 1}{1 \over n + n}} & \ds{<} &
\ds{\sum_{k = 0}^{n - 1}{1 \over k + n}} & \ds{<} &
\ds{\sum_{k = 0}^{n - 1}{1 \over 0 + n}}
\\[2mm]
\ds{\half} & \ds{<} &
\ds{\sum_{k = 0}^{n - 1}{1 \over k + n}} & \ds{<} &
\ds{1}
\\[2mm]
\ds{1 \over 2n} & \ds{<} &
\ds{{1 \over n}\sum_{k = 0}^{n - 1}{1 \over k + n}} & \ds{<} &
\ds{1 \over n}
\end{array}
\end{align}
A: 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 $n\geq3$.
A: 
We show the inequality chain 
  \begin{align*}
\frac{1}{2}\leq\sum_{k=0}^n\frac{1}{k+n}\leq 1\qquad\qquad\qquad n\geq 1\tag{1}
\end{align*}
is not valid for $n=1,2$ and valid for $n\geq 3$.

We denote the sum with $A(n):=\sum_{k=0}^n\frac{1}{k+n}$.

Case $n=1,2,3$ :
\begin{align*}
A(1)&=\sum_{k=0}^1\frac{1}{k+1}=1+\frac{1}{2}=\frac{3}{2}>1\\
A(2)&=\sum_{k=0}^2\frac{1}{k+2}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}>1\\
A(3)&=\sum_{k=0}^3\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{19}{20}<1\\
\end{align*}
We observe $A(1)$ and $A(2)$ are greater than $1$, while $\frac{1}{2}\leq A(3)\leq 1$.
Conclusion:
  
  
*
  
*The inequality chain (1) is not valid for $n=1,2$.
  
*Since $\frac{1}{2}\leq A(3)=\frac{19}{20}\leq 1$ the inequality chain (1) is valid for $n=3$.

$$ $$

Monotonicity of $A(n)$:
We want to compare $A(n)$ with $A(n+1)$. We obtain for $n\geq 1$
\begin{align*}
A(n+1)&=\sum_{k=0}^{n+1}\frac{1}{k+n+1}\\
&=\sum_{k=1}^{n+2}\frac{1}{n+k}\\
&=\sum_{k=0}^{n}\frac{1}{k+n}+\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n}\\
&=A(n)+\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n}
\end{align*}
When we consider with some help of Wolfram Alpha the function
  $$f(x)=\frac{1}{2x+1}+\frac{1}{2x+2}-\frac{1}{x}$$
  with $x$ real, we see there is just one zero at $x=-\frac{2}{3}$. Since $f(1)=-\frac{5}{12}$, the function is  negative for $x\geq 1$ and so
\begin{align*}
A(n+1)-A(n)=\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n}<0\qquad\qquad n\geq 1
\end{align*}
Conclusion:
  
  
*
  
*$A(n)$ is monotonically decreasing with  increasing $n$.
  
*Since $A(3)\leq 1$ we see that $1$ is an upper limit of $A(n)$ for $n\geq 3$.

Finally we  show  $\frac{1}{2}$ is a lower limit of $A(n)$.

Harmonic numbers $H_n$:
Note that $A(n)$ is closely related with harmonic numbers $H_n=\sum_{k=1}^n\frac{1}{k}$.
We obtain
  \begin{align*}
A(n)=\sum_{k=0}^n\frac{1}{k+n}=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^{n-1}\frac{1}{k}=H_{2n}-H_{n-1}\qquad\qquad n\geq 1
\end{align*}
The harmonic numbers are asymptotically equal to
  \begin{align*}
H_n\sim \ln n+\gamma
\end{align*}
  with $\gamma$ the Euler constant. We obtain
  \begin{align*}
\lim_{n\rightarrow\infty}A(n)&=\lim_{n\rightarrow \infty}\left(H_{2n}-H_{n-1}\right)\\
&\sim
\ln(2n)+\gamma-\ln(n-1)-\gamma\\
&\sim\ln 2
\end{align*}
Conclusion:
  
  
*
  
*Since $\ln 2\doteq 0.69314>\frac{1}{2}$ we see $A(n)\geq\frac{1}{2}$ for all $n\geq 3$.
  

$$ $$

Summary:
  
  
*
  
*The inequality chain (1) is not valid for $n=1,2$ and valid for all $n\geq 3$.
  
*The sum is monotonically decreasing with increasing $n$.
  $$\sum_{k=0}^n\frac{1}{k+n}\searrow$$
  
*The limit of the sum is  $\ln 2$. 
$$\lim_{n\rightarrow\infty}\sum_{k=0}^n\frac{1}{k+n}=\ln 2$$

A: It holds when $k=1$
$$\frac { 1 }{ n+1 } +\frac { 1 }{ n+2 } +...+\frac { 1 }{ 2n } \ge \overset { n }{ \overbrace { \frac { 1 }{ 2n } +\frac { 1 }{ 2n } +...\frac { 1 }{ 2n }  }  } =n\frac { 1 }{ 2n } =\frac { 1 }{ 2 } \\ \\ \frac { 1 }{ n+1 } +\frac { 1 }{ n+2 } +...+\frac { 1 }{ 2n } \le \overset { n }{ \overbrace { \frac { 1 }{ n+1 } +\frac { 1 }{ n+1 } +...+\frac { 1 }{ n+1 }  }  } =\frac { n }{ n+1 } \le 1\\ $$
