Proof that a sequence of positive numbers satisfies $P_n ^2 \leq P_{n-1}\,P_{n+1}$.

Question

Let $(P_n)$ be a sequence such that $P_0 = 1$ and $P_n = (\log{(n+1)})^n$.

I'm trying to prove that $P_n ^2 \leq P_{n-1}\,P_{n+1}$ for $n \geq 1$ using an induction argument.

For $n=1$ we have that $(\log 2)^2 = 0.48$ and $1\cdot (\log 3)^2 = 1.206$, hence $P_1 ^{2} \leq P_{0}\,P_2$.

Given $n > 1$, suppose that $P_n ^2 \leq P_{n-1}\,P_{n+1}$, i.e $$ (\log (n+1))^{2n} \leq (\log (n))^{n-1}\, (\log (n+2))^{n+1} $$

we want to show that

$$ (\log (n+2))^{2(n+1)} \leq (\log (n+1))^{n}\, (\log (n+3))^{n+2}.$$

However, I've been trying here without any success.

Help?

Answer 1

Let $f(x)=x \log\log(x+1)$. Then $$f'(x)=\log \log(x+1)+\dfrac{x}{(x+1)\log(x+1)},$$ $$f''(x)=\dfrac{1}{(x+1)\log(x+1)}+\dfrac{1}{(x+1)^2\log(x+1)}-\dfrac{x}{(x+1)^2\log^2(x+1)}=\dfrac{1}{(x+1)\log(x+1)}\left( 1+\dfrac{1}{x+1}-\dfrac{1}{(x+1)\log(x+1)}\right )\geq 0.$$ So $y=f(x)$ is concave for $x\geq 0$. In particular $$f(n) \leq \dfrac{f(n-1)+f(n+1)}{2}.$$ So $$n\log \log(n+1)\leq \dfrac{(n-1)\log \log n + (n+1)\log \log(n+2)}{2}$$ which implies the inequality after exponentiating both sides.

Answer 2

One (probably long) way is $$P_n ^2 \leq P_{n-1}\,P_{n+1} \iff \frac{P_n}{P_{n-1}} \leq \frac{P_{n+1}}{P_n}$$ which is equivalent to showing that the sequence $a_n=\frac{P_n}{P_{n-1}}$ is increasing. For this purpose, let's look at the function $$\color{blue}{f(x)=\ln\left(\frac{(\ln(x+1))^x}{(\ln(x))^{x-1}}\right)} =\ln(\ln(x+1))+(x-1)\ln\left(\frac{\ln(x+1)}{\ln(x)}\right) \tag{1}$$ with $$\color{red}{f'(x)}=\frac{x}{(1+x)\ln(1+x)}+\ln(\ln(1+x))-\frac{x-1}{x\ln(x)}-\ln(\ln(x))=\\ \color{red}{g(x)-g(x-1)} \tag{2}$$ where $$g(x)=\frac{x}{(1+x)\ln(1+x)}+\ln(\ln(1+x))$$ with $$g'(x)=\frac{(x+2)\ln(1+x)-x}{(1+x)^2\ln^2(1+x)}\geq0, \forall x\geq0$$ and $$g(x)\geq g(x-1), \forall x\geq 1 \overset{\color{red}{(2)}}{\Rightarrow} f'(x) \geq 0$$ of $f(x)$ is ascending function $\forall x\geq 1$. But then $e^{f(x)}$ is also ascending $\forall x\geq 1$, where $$e^{f(x)}\overset{\color{blue}{(1)}}{=}\frac{(\ln(x+1))^x}{(\ln(x))^{x-1}}$$ and $$a_n=e^{f(n)}\leq e^{f(n+1)}=a_{n+1}$$