I am currently trying to exercise for a competition and have the following problem for it: $A_n = \log_{n}(n+1)+\log_{n+1}(n)$
where $n$ is natural number. I must prove that $A_n$ can never be a whole number and I must calculate the whole part of $A_n$ . I also must study the monotonicity of the $A_n$ sequence.
I tried by converting the logarithms to the natural base but failed with it.
I would gladly appreciate your help in solving this problem, I am pretty stuck with it. Thanks in advance!
$\begin{array}\\ A_n &= \log_{n}(n+1)+\log_{n+1}(n)\\ &= \dfrac{\log(n+1)}{\log(n)}+\dfrac{\log(n)}{\log(n+1)}\\ &=x+\dfrac1{x} \qquad x=\dfrac{\log(n+1)}{\log(n)} \gt 1\\ &=(\sqrt{x}-\dfrac1{\sqrt{x}})^2+2\\ &\gt 2\\ \end{array} $
If $A_n \ge 3$ then $\sqrt{x}-\dfrac1{\sqrt{x}} \ge 1$.
Let $y=\sqrt{x} \gt 1$.
If $y-\dfrac1{y} \ge 1$ then $y^2-y \ge 1$ or $(y-\frac12)^2 \ge \frac54 $ or $y \ge \frac{1+\sqrt{5}}{2} $ so $x =y^2 \ge \frac{1+2\sqrt{5}+5}{4} = \frac{3+\sqrt{5}}{2} \gt 2.618 $.
But, using natural logs,
$\begin{array}\\ x &=\dfrac{\log(n+1)}{\log(n)}\\ &=\dfrac{\log(n)+\log(1+1/n)}{\log(n)}\\ &=1+\dfrac{\log(1+1/n)}{\log(n)}\\ &\lt 1+\dfrac1{n\log(n)} \qquad \log(1+z)<z\\ &\lt 1+\dfrac1{2\log(2)} \qquad (z\log(z))'=1+\log(z)>1 \text{ for } z>1\\ &\lt 1.8\\ \end{array} $
which contradicts $x > 2.618$.
Therefore $2 < A_n < 3$.