Prove without differentiation that $\log_{n+1}n$ is increasing 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?
 A: 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 $$\log_{n+1}\left(n\right)<\log_{n+2}\left(n+1\right)\Leftrightarrow\frac{\log\left(n\right)}{\log\left(n+1\right)}<\frac{\log\left(n+1\right)}{\log\left(n+2\right)}\Leftrightarrow\log\left(n\right)\log\left(n+2\right)<\log^{2}\left(n+1\right)$$
A: 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.
A: 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 $\sup_n x_n = 1$.
