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Is it possible to prove the monotonicity of $(a_n)_n =\frac{\ln n}{\sqrt n}$ without using derivatives?

MJD
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zar
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  • Prove that $f(x)=\ln x/x$ is decreasing for positive real $x$ first. Then $a_n=2f(\sqrt n)$ and $\sqrt n$ is increasing and $a_n$ is therefore decreaasing. – Thomas Andrews Oct 10 '12 at 17:15
  • Without derivatives? – zar Oct 10 '12 at 17:32
  • It's not clear why you write $(a_n)_n$. Usually, we'd just write $a_n=\cdots$ – Thomas Andrews Oct 10 '12 at 17:32
  • It should be $(a_n)_{n\in\mathbf{N}}$, but I can edit it. – zar Oct 10 '12 at 17:33
  • I didn't mean that I knew how to show $\ln x/x$ is decreasing, just that it might be an easier problem to solve than $\ln x/\sqrt x$. – Thomas Andrews Oct 10 '12 at 18:08
  • Also, it isn't strictly monotonic, since $a_1=0$ and the other values are positive. – Thomas Andrews Oct 10 '12 at 18:22
  • Yes, I know that, I would like to prove the definite (? definitive? sorry for my english) monotonicity – zar Oct 10 '12 at 19:40
  • as Thomas said, we need to prove $\frac{ln n}{n} >\frac{ln n+1}{n+1} $ i.e. $n^{\frac{1}{n}} > {n+1}^{\frac{1}{n+1}}$, $n^{n+1} > {n+1}^n$, $ n > {1+ \frac{1}{n}}^n$. right side tends to e as n tends to infiniy. – alekhine Oct 10 '12 at 20:21
  • If we want to prove that $\ln n/\sqrt{n}$ is decreasing, we should prove that $\ln x/x$ is decreasing for $x\in\mathbf{R}$. It is not sufficient to prove it monotonicity for $n\in\mathbf{N}$ – zar Oct 11 '12 at 12:14

2 Answers2

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The sequence $\left\{\dfrac{\ln{n}}{n}\right\}_{n \in \mathbb{N}, \;n \geqslant{3}}$ is monotonically decreasing: $$\dfrac{\ln{n}}{n}-\dfrac{\ln{(n+1)}}{n+1}=\dfrac{(n+1)\ln{n}-n\ln{(n+1)}}{n(n+1)}= \dfrac{\ln{n^{n+1}}-\ln{(n+1)^n}}{n(n+1)}=\dfrac{\ln\dfrac{n^{n+1}}{(n+1)^n}}{n(n+1)}=\dfrac{-\ln\dfrac{(n+1)^n }{n^{n+1}}}{n(n+1)}=-\dfrac{\ln\left[\left(1+\dfrac{1 }{n}\right)^n \cdot\dfrac{1}{n}\right]}{n(n+1)}>0$$ for $n>3.$
The last inequality follows from the well-known estimate $${2} \leqslant \left(1+\dfrac{1 }{n}\right)^n \leqslant {3} \quad \forall n \in \mathbb{N},$$ so that for $n>3$ $$\left(1+\dfrac{1 }{n}\right)^n \cdot\dfrac{1}{n}<1.$$

M. Strochyk
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For $a_n=\dfrac{\ln{n}}{\sqrt{n}}$ we denote $f_n=\dfrac{1}{a_n^2}=\dfrac{n}{\ln^2{n}}.$ In order to prove that $\left\lbrace a_n \right\rbrace $ is decreasing, we consider \begin{gather*} f_{n+1}-f_n=\dfrac{n+1}{\ln^2(n+1)}- \dfrac{n}{\ln^2{n}}= \dfrac{(n+1)\ln^2{n}-n\ln^2(n+1)}{\ln^2{n} \ \ln^2(n+1)}=\dfrac{A}{B} \end{gather*} and show that sequence $\left\lbrace f_n \right\rbrace$ increases: $f_{n+1}-f_n >0 \quad \left(\forall n \geqslant n_0 \right).$ Numerator $A$ may be rewritten as \begin{gather*} A=(n+1)\ln^2{n}-n\ln^2(n+1)=(n+1)\ln^2{n}-n\ln^2\left(n\left(1+\dfrac{1}{n}\right)\right)=\\ (n+1)\ln^2{n}-n\left(\ln{n}+\ln\left(1+\dfrac{1}{n}\right)\right)^2=\\ n \ln^2{n}+\ln^2{n}-n \ln^2{n}-2n \ln{n}\ln\left(1+\dfrac{1}{n}\right)-n \ln^2\left(1+\dfrac{1}{n}\right)=\\ \ln^2{n}-n \ln{n}\ln\left(1+\dfrac{1}{n}\right)-n \ln{n}\ln\left(1+\dfrac{1}{n}\right)-n \ln^2\left(1+\dfrac{1}{n}\right)=\\ \ln{n}\left(\ln{n}-n\ln\left(1+\dfrac{1}{n}\right)\right)-n\ln\left(1+\dfrac{1}{n}\right)\left( \ln{n}+\ln\left(1+\dfrac{1}{n}\right)\right)=\\ \ln{n}\left(\ln{n}-\ln\left(1+\dfrac{1}{n}\right)^n\right)-\ln\left(1+\dfrac{1}{n}\right)^n\left( \ln{n}+\ln\left(1+\dfrac{1}{n}\right)\right). \end{gather*} Next, applying the estimate $${2} \leqslant \left(1+\dfrac{1 }{n}\right)^n \leqslant {3} \quad \left(\forall n \in \mathbb{N}\right),$$ we obtain \begin{gather} \ln{n}\left(\ln{n}-\ln\left(1+\dfrac{1}{n}\right)^n\right) \geqslant \ln{n}\left(\ln{n}-\ln{3} \right);\\ -\ln\left(1+\dfrac{1}{n}\right)^n\left( \ln{n}+\ln\left(1+\dfrac{1}{n}\right)\right) \geqslant -\ln{3}(\ln{n}+\ln{2}). \end{gather} Thus, by adding these inequalities, $$ A \geqslant {\ln{n}\left(\ln{n}-\ln{3} \right) -\ln{3}(\ln{n}+\ln{2})}= \ln^2{n}-2\ln{3} \ \ln{n}-\ln{2} \ \ln{3}.$$ Quadratic polynomial $ x^2-2\ln{3}\cdot x-\ln{2}\ln{3}>0 $ for $x\in\left(-\infty,\, x_1\right)\cup{\left(x_2, \,+\infty\right)}, $ where \begin{gather*}x_1=\ln{3}- \sqrt{\ln^2{3}+\ln{2}\ln{3}}, \\ x_2=\ln{3}+ \sqrt{\ln^2{3}+\ln{2}\ln{3}}. \end{gather*} Thus, $A>0$ for $\ln{n}>\ln{3}+ \sqrt{\ln^2{3}+\ln{2}\ln{3}}\approx 2.501626534,$ respectively $n\geqslant n_0= \left \lfloor{e^{2.501626534}}\right \rfloor+1=\left\lfloor {12.20232533}\right\rfloor+1=13.$ This value is less precise than exact value $e^2,$ which is point of minimum for $f(x)=\dfrac{x}{\ln^2{x}}$.

M. Strochyk
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