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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....

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    $\begingroup$ could you show what/how you tried? $\endgroup$ – Alex Sep 2 '16 at 19:44
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    $\begingroup$ The inequality is false as stated. You can fix it by starting the summation at $k=1$. $\endgroup$ – grand_chat Sep 2 '16 at 20:52
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    $\begingroup$ As stated the inequality only holds for $n\geq 3$ $\endgroup$ – Winther Sep 2 '16 at 21:10
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    $\begingroup$ You are simply taking arithmetical mean of several elements, the largest of them is $\frac1n$ and the smallest one is $\frac1{2n}$. $\endgroup$ – Martin Sleziak Sep 2 '16 at 22:29
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    $\begingroup$ @MartinSleziak No, there are $n+1$ terms, not $n$ (but my guess is that the OP simply miscopied their homework, starting their sums at $\frac1n$ instead of $\frac1{n+1}$). (As user grand_chat already suggested.) $\endgroup$ – Did Sep 3 '16 at 11:25
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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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  • $\begingroup$ it's not hint it's answer :) thanks very much.... $\endgroup$ – Kamikaze Sep 2 '16 at 19:49
  • $\begingroup$ you're welcome) $\endgroup$ – haqnatural Sep 2 '16 at 19:51
  • $\begingroup$ As written, this is incorrect. There are $n+1$ terms in the sum, not $n$. But I think it can be corrected. $\endgroup$ – MPW Sep 2 '16 at 20:04
  • $\begingroup$ thanks for observation @MPW $\endgroup$ – haqnatural Sep 2 '16 at 20:06
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    $\begingroup$ @MPW The inequality holds if the sum starts at $k=1$, but the inequality is wrong if it starts at $k=0$ (If $n=1$, the incorrect inequality would assert $\frac12\le \frac11 +\frac12\le 1$). $\endgroup$ – grand_chat Sep 2 '16 at 20:50
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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}

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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 $n\geq3$.

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    $\begingroup$ It is in any case a very nice solution! (+1) $\endgroup$ – Markus Scheuer Sep 3 '16 at 11:08
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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 $n=3$.

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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 $n\geq 3$.

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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 $\ln 2$.

    $$\lim_{n\rightarrow\infty}\sum_{k=0}^n\frac{1}{k+n}=\ln 2$$

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  • $\begingroup$ wow thank you very much for interesting. $\endgroup$ – Kamikaze Sep 3 '16 at 13:29
  • $\begingroup$ @Kamikaze: You're welcome! I've updated the summary to better see the additional information of this approach. $\endgroup$ – Markus Scheuer Sep 3 '16 at 13:37

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