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I found an inequality: $$\ln\left(\frac{n+1}{n}\right)\le\frac{2}{n+1},\forall n\in\mathbb{N}^+.$$

I tried induction. It is obvious if $k=1$, when $n=k$, $\ln\sqrt{\left(\frac{k+1}{k}\right)^{k+1}}\le 1$, but bogged down for $n=k+1$:

$$\ln\sqrt{\left(\frac{k+2}{k+1}\right)^{k+2}}\le 1$$

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  • $\begingroup$ That follows easily from the "well-known" inequality $\ln x \le x - 1$ for $x > 0$. $\endgroup$ – Martin R Jul 20 '17 at 9:26
  • $\begingroup$ Shouldn't it be $\ln\sqrt{\left(\frac{k+1}{k}\right)^{k+1}}$? $\endgroup$ – Arthur Jul 20 '17 at 9:26
  • $\begingroup$ Yes, thanks. I corrected it. $\endgroup$ – MathArt Jul 20 '17 at 10:33
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Take $x=\frac1n$ so you will have $$\ln(1+x)\leq \frac{2x}{x+1}$$ now suppose $f(x)=\ln(1+x)- \frac{2x}{x+1}\\x \in [0,1]$ $$f'(x)=\frac{1}{x+1}-\frac{2}{(x+1)^2}=\frac{1+x-2}{(x+1)^2} \to x=1$$ so $f'(x) \leq 0 \text { when x is } \in [0,1]$ so $f(x)$ is decreasing in this interval so $$f(x)\leq f(0) \text{ when } x\in [0,1]\\ \to f(x)\leq f(0)\\f(x) \leq \ln(1+0)-0\\f(x)\leq 0\\\ln(1+x)- \frac{2x}{x+1}\leq 0\\ \ln(1+x)\leq \frac{2x}{x+1} , x \in [0,1] \\and \\\ln(1+x)\leq \frac{2x}{x+1} , x \in (0,1]$$ now put $x=\frac1n$

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    $\begingroup$ Should be $f'(x)\le0$? $\endgroup$ – MathArt Jul 20 '17 at 10:26
  • $\begingroup$ @MathArt :Yes, you are right ,I fixed it . $\endgroup$ – Khosrotash Jul 20 '17 at 10:57
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$$\ln((n+1)/n) = \ln(n+1)-\ln(n) = \int_n^{n+1}\frac{1}{x} dx \le \frac{1}{n} \le \frac{2}{n+1}$$

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Apply $\exp$ (which is strictly increasing) to both sides. Then your inequality is equivalent to $$ \frac{n+1}{n}\le e^{\frac2{n+1}}$$ One of the most useful inequalities about the exponential is $$ e^t\ge 1+t.$$ This gives us $$ e^{\frac2{n+1}}\ge 1+\frac2{n+1}\stackrel{(*)}\ge1+\frac1n=\frac{n+1}n$$ where $(*)$ follows from $\frac2{n+1}-\frac1n=\frac{2n-(n+1)}{n(n+1)}\ge0$.

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HINT

Use the fact that $(1 + \frac 1 n)^n \lt e$ See Euler's constant

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  • $\begingroup$ Thanks, maybe it is better to show $n\ln(1 + \frac 1 n) \le 1$ and $\ln(1+\frac{1}{n})\lt 1$ and add them. But seems the equality is not smooth? $\endgroup$ – MathArt Jul 20 '17 at 10:48
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Start from:

1) ln$ (1 +x) \le x$, for $x \ge 0$.

2) $ n +1 \le 2n$,

$\rightarrow 1/n \le 2/(n+1)$.

Combining :

ln$ (1+ 1/n) \le 1/n \le 2/(n +1)$,

where $n$ is a positive integer.

Proof of 1):

Series expansion for $e^x$:

$ e^x = 1+ x + x^2/2! + x^3/3! +.... $

$\rightarrow$ $(1+x) \le e^x$ for $x \ge 0$.

ln is a stricly monotonously increasing function , hence

ln $(1+ x) \le x$ for $x\ge0$.

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Integrate $\frac{1}{x}$ between $n$ and $n+1$: $\ln \frac{n+1}{n} = \ln (n+1) - \ln (n) = \displaystyle\int_n^{n+1} \frac{1}{x} \operatorname{d}x$. As $x\to \frac{1}{x}$ is decreasing, you get $\ln \frac{n+1}{n}\leq \displaystyle\int_n^{n+1} \frac{1}{n} \operatorname{d}x = \frac{1}{n}$.

It is moreover clear that $\frac{1}{n} \leq \frac{2}{n+1}$ for $n\geq 1$, so you get $\ln \frac{n+1}{n} \leq \frac{2}{n+1}$

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Your question is equivalent to prove the following form $$ (2-\frac{n-1}{n})\leq 2^{(1-\frac{n-1}{n+1})} \tag{1} $$ The relation $(1)$ is true becuse $\frac{n-1}{n} \geq \frac{n-1}{n+1}$.

Edit: If the relation $(1)$ be true then we can conclude that: $$ \left\{ \begin{array}{c} (2-\frac{n-1}{n})\leq 2^{(1-\frac{n-1}{n+1})} \\ \\ 2^{(1-\frac{n-1}{n+1})} \leq e^{(1-\frac{n-1}{n+1})} \end{array} \right. \Rightarrow (2-\frac{n-1}{n})\leq e^{(1-\frac{n-1}{n+1})} \tag{2} $$ $$ \ln (2-\frac{n-1}{n})\leq (1-\frac{n-1}{n+1}) \Rightarrow \ln\left(\frac{n+1}{n}\right)\le\frac{2}{n+1} $$

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  • $\begingroup$ I have no idea about the equivalence? $\endgroup$ – MathArt Jul 20 '17 at 10:55
  • $\begingroup$ @MathArt I edit my question. $\endgroup$ – Amin235 Jul 20 '17 at 13:10

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