I want to show that the sequence $\{ \log_{n+1}n \}$ is increasing without differentiation.
I don't have any idea. How can I prove that?
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Do you allow concavity of the natural log? – a25bedc5-3d09-41b8-82fb-ea6c353d75ae Jul 23 '16 at 11:10
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Did you try to use the identity $$\log_{n+1} n = \frac{\log n}{\log (n+1)}?$$ – Siminore Jul 23 '16 at 11:10
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1@user4759923 You mean, the concavity of $\log\log$ the iterated natural logarithm? – Did Jul 23 '16 at 11:12
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Yes, I wanted to use the identity and concavity of the iterated log – a25bedc5-3d09-41b8-82fb-ea6c353d75ae Jul 23 '16 at 11:15
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@Siminore Yes. But it didn't help me. How to prove ${ \log(n+2) }^2 > \log (n) \log(n+2)$? I tried to use Jensen's inequality, but it also failed. – PPPiRi Jul 23 '16 at 11:19
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@user4759923 concavity of iterated log is not allowed. I mean, if you cannot show that iterated log is concave without differentiation. – PPPiRi Jul 23 '16 at 11:20
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Why is differentiation not allowed? – a25bedc5-3d09-41b8-82fb-ea6c353d75ae Jul 23 '16 at 11:24
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I want to explain the proof to student who don't know about differentiation. (and because this restriction inspires my curiosity.) – PPPiRi Jul 23 '16 at 11:30
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Although not a proof, you could just explain (using the identity) that when $n=1$, the value is $0$, when $n>1$, the value is positive and less than one, and that, as $n\to \infty$, the 1 in the denominator becomes unimportant so that $\log_{n+1}n\to 1$. – smcc Jul 23 '16 at 11:39
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You can use the AM-GM inequality. We have $$\color{blue}{\log\left(n\right)\log\left(n+2\right)}\leq\frac{\left(\log\left(n\right)+\log\left(n+2\right)\right)^{2}}{4}$$ $$=\frac{\log^{2}\left(n^{2}+2n\right)}{4}<\frac{\log^{2}\left(n^{2}+2n+1\right)}{4}=\color{red}{\log^{2}\left(n+1\right)}$$ and this conclude the proof since $$\log_{n+1}\left(n\right)<\log_{n+2}\left(n+1\right)\Leftrightarrow\frac{\log\left(n\right)}{\log\left(n+1\right)}<\frac{\log\left(n+1\right)}{\log\left(n+2\right)}\Leftrightarrow\log\left(n\right)\log\left(n+2\right)<\log^{2}\left(n+1\right)$$
Marco Cantarini
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We use only these properties of the logarithm: $$ \begin{align} \log_a(b)&=\frac{\ln b}{\ln a},\\ \ln(xy)&=\ln(x) +\ln(y)&\text{for }x,y>0,\\ \ln(x)&>0&\text{for }x>1,\\ \ln(x)&<0&\text{for }0<x<1.\end{align}$$ With these we find $$\begin{align}\ln(n)\ln(n+2)&=\bigl(\ln(n+1)+\ln(1-\tfrac1{n+1})\bigr)\cdot\bigl(\ln(n+1)+\ln(1+\tfrac1{n+1})\bigr)\\ &=\ln^2(n+1)\\&\quad +\ln(n+1)\cdot\left(\ln(1-\tfrac1{n+1})+\ln(1+\tfrac1{n+1})\right)\\ &\quad +{\ln(1-\tfrac1{n+1})}\cdot{\ln(1+\tfrac1{n+1})}\\ &=\ln^2(n+1)\\&\quad +\underbrace{\ln(n+1)}_{>0}\cdot\underbrace{\ln\bigl(1-\tfrac1{(n+1)^2}\bigr)}_{<0}\\ &\quad +\underbrace{\ln(1-\tfrac1{n+1})}_{<0}\cdot\underbrace{\ln(1+\tfrac1{n+1})}_{>0}\\ &<\ln^2(n+1).\end{align}$$
Thus with $a_n:=\log_{n+1}(n)=\frac{\ln(n)}{\ln(n+1)}$, we have $$\begin{align}a_{n+1}-a_n &= \frac{\ln(n+1)}{\ln(n+2)}-\frac{\ln(n)}{\ln(n+1)}\\ &=\frac{\ln^2(n+1)-\ln(n)\ln(n+2)}{\ln(n+2)\ln(n+1)}\\ &>0 \end{align}$$ as desired.
Hagen von Eitzen
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From \begin{align} x_n = \log_{n+1} n &\iff (n+1)^{x_n} =n\\&\iff x_n\log(n+1)=\log n\\&\iff x_n = \frac{\log n}{\log(n+1)} \end{align} and $$\log(n+1) = \log\left(n\left(1+\frac1n\right)\right) = \log n + \log\left(1+\frac1n\right) $$ we have $$x_n = \frac{\log n}{\log n + \log\left(1+\frac1n\right)} = \left(1+\frac{\log\left(1+\frac1n\right)}{\log n}\right)^{-1}. $$ Since $x\mapsto \log x$ is an increasing function, it follows that $x_n<x_{n+1}$.
Moreover, $$\lim_{n\to\infty}\log\left(1+\frac1n\right)=0,$$ so the sequence converges to $\sup_n x_n = 1$.
Math1000
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What you have written so far seems misleading: you have $\frac{\log(n)}{\log(n)+\log(1+1/n)}$, you note that $\log(1+1/n)$ is decreasing and so you conclude that the whole thing is increasing. But that's not enough, because the other term in the denominator is increasing. Indeed, if you factor a $\log(n)$ out of numerator and denominator, you find that the problem basically reduces right back to the original problem: you need $\frac{1}{1+\frac{\log(1+1/n)}{\log(n)}}$ to be increasing so you need $\log(1+1/n)/\log(n)$ to be decreasing. – Ian Jul 23 '16 at 11:46
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1Er, yes, sorry, you're right, that much is evident (I thought it wound up being just the reciprocal of the thing you want to show is increasing for a moment). Anyway, you should still include this additional algebraic step; as it stands it sounds like the situation is much more general than it is. – Ian Jul 23 '16 at 11:51
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1You now actually have a typo, there should be no $\log(n)$ on the outside of the right side of the second to last equation. It was canceled with the factor of $\log(n)$ that you factored. – Ian Jul 23 '16 at 14:25
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