Proving that ${(1-\frac{2}{x^2})}^x < \frac{x-1}{x+1}$ for any $x > 2$.

Question

Proof that ${\left(1-\dfrac{2}{x^2}\right)}^x\!< \dfrac{x-1}{x+1}$ for any $x > 2$.

any ideas?

Answer 1

Hi @max We can use Binomial series as we have :

$$\left(1-\frac{2}{x^{2}}\right)\leq \left(\frac{\left(x-1\right)}{x+1}\right)^{\frac{1}{x}}=\left(\frac{\left(x-1\right)}{x+1}\right)^{1+\frac{1}{x}-1}$$

So at order two we have :

$$\left(\frac{\left(x-1\right)}{x+1}\right)^{-1}\left(1+\left(\frac{\left(x-1\right)}{x+1}-1\right)\left(1+\frac{1}{x}\right)+\frac{1}{2}\left(\frac{\left(x-1\right)}{x+1}-1\right)^{2}\left(1+\frac{1}{x}\right)\left(\frac{1}{x}\right)\right)\leq \left(\frac{\left(x-1\right)}{x+1}\right)^{\frac{1}{x}}$$

But :

$$\left(\frac{\left(x-1\right)}{x+1}\right)^{-1}\left(1+\left(\frac{\left(x-1\right)}{x+1}-1\right)\left(1+\frac{1}{x}\right)+\frac{1}{2}\left(\frac{\left(x-1\right)}{x+1}-1\right)^{2}\left(1+\frac{1}{x}\right)\left(\frac{1}{x}\right)\right)=\left(1-\frac{2}{x^{2}}\right)$$

We are done !

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Some idea fo someone who wants to show Angelo problem :

For $x\geq 2$ we have :

$$\cos\left(\frac{2}{x}\right)-\cos\left(\frac{2}{\sqrt{x}}\right)\geq 2\left|\frac{x-1}{x^{2}+\frac{2}{3}+\frac{1}{3}x}\right|\geq \left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}-\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)^{\frac{1}{\sqrt{x}}}$$

Remains to show :

$$\lim_{n\to \infty}\cos\left(\frac{2}{x^{2n}}\right)-\left(\frac{x^{2n}-1}{x^{2n}+1}\right)^{\frac{1}{x^{2n}}}$$

Answer 2

Taking logarithms leads to

$$ x\log\left(1-\frac2{x^2}\right)\lt\log(x-1)-\log(x+1) $$

and thus

$$ \log\left(1-\frac2{x^2}\right)\lt\frac1x\left(\log\left(1-\frac1x\right)-\log\left(1+\frac1x\right)\right)\;. $$

So we want to show that

$$ \log\left(1-2z^2\right)\lt z(\log(1-z)-\log(1+z)) $$

for $0\lt z\lt\frac12$. Expanding both sides into power series leads to

$$ -\sum_{k=1}^\infty\frac{\left(2z^2\right)^k}k\lt-z\left(\sum_{j=1}^\infty\frac{z^j}j-\sum_{j=1}^\infty\frac{(-z)^j}j\right) $$

and thus

$$ \sum_{k=1}^\infty\frac{2^k}kz^{2k}\gt\sum_{k=1}^\infty\frac2{2k-1}z^{2k}\;. $$

The claim follows since the terms for $k=1$ are equal and all other terms are larger on the left. This in fact holds for $0\lt z\lt\frac1{\sqrt2}$ and thus for $x\gt\sqrt2$.

Answer 3

Let $f(x)=\ln\frac{x-1}{x+1}-x\ln(1-2/x^2)$. It suffices to show that $f(x)>0$ for $x>2$, as $\ln$ is an increasing function. Using the series for $\ln(1+a)=a-a^2/3+a^3/3-a^4/4+\dots$, we get the asymptotic expansion $$ \ln \frac{x-1}{x+1}=\ln \left(1-\frac1x\right)-\ln \left(1+\frac1x\right)=-\frac{2}{x}-\frac{2}{3x^3}-\frac{2}{5x^5}-\frac{2}{7x^7}-\dots $$ and $$ -x\ln\left(1-\frac2{x^2}\right)=\frac2x+\frac{2^2}2 \frac{1}{x^3}+\frac{2^3}3 \frac{1}{x^5}+\frac{2^4}4 \frac{1}{x^7}+\dots $$ hence $$ f(x)=\sum_{k=1}^\infty \frac 1{x^{2k+1}}\left(\frac{2^{k + 1}}{k + 1} - \frac2{2k+1}\right)>0 $$ since $$ \frac{2^{k + 1}}{k + 1} - \frac2{2k+1}=2\frac{2^k(2k+1)-(k+1)}{(k+1)(2k+1)}>0. $$ (Note that the inequality $f(x)>0$ holds for all $x>\sqrt 2$.)