# Prove $0.9999^{\!101}<0.99<0.9999^{\!100}$

Prove $$0.9999^{\!101}<0.99<0.9999^{\!100}$$ I think its original idea is $$(1-x)^{(1-\frac{1}{x})}<(1+x)^{(\frac{1}{x})} \tag{I can't prove!}$$ For $$x=100^{\!-1}\,\therefore\,(1-x)^{(1-\frac{1}{x})}<(1+x)^{(\frac{1}{x})}\,\therefore\,0.99^{\!-99}<1.01^{\!100}$$.

Furthermore $$0.99^{\!-99}\times0.99^{\!100}=0.99<1.01^{\!100}\times0.99^{\!100}=0.9999^{\!100}$$.

For $$x=-100^{\!-1}\,\therefore\,(1-x)^{(1-\frac{1}{x})}<(1+x)^{(\frac{1}{x})}\,\therefore\,1.01^{\!101}<0.99^{\!-100}$$.

Furthermore $$1.01^{\!101}\times0.99^{\!101}=0.9999^{\!101}<0.99^{\!-100}\times0.99^{\!101}=0.99$$.

• It's not too hard to prove this for the two exponential parts by taking the log of each side. I'm curious if a similar approach can be used for the middle.
– Max
May 28 '19 at 7:06

$$0.99<0.9999^{100}$$ it's $$1-\frac{1}{100}<\left(1-\frac{1}{100^2}\right)^{100},$$ which is true by Bernoulli: $$\left(1-\frac{1}{100^2}\right)^{100}>1-100\cdot\frac{1}{100^2}=1-\frac{1}{100}.$$ $$0.9999^{101}<0.99$$ it's $$\left(1-\frac{1}{100^2}\right)^{101}<1-\frac{1}{100}$$ or $$\left(1-\frac{1}{100^2}\right)^{100}\left(1+\frac{1}{100}\right)<1$$ or $$\left(1-\frac{1}{100^2}\right)^{100}<1-\frac{1}{101},$$ which is true because $$\left(1-\frac{1}{100^2}\right)^{100}<1-\binom{100}{1}\cdot\frac{1}{100^2}+\binom{100}{2}\cdot\frac{1}{100^4}<1-\frac{1}{101}.$$ I used the following lemma.

Let $$0 and $$n>2.$$ Prove that: $$(1-x)^n<1-nx+\frac{n(n-1)}{2}x^2.$$

Proof.

We need to prove that $$f(x)>0,$$ where $$f(x)=1-nx+\frac{n(n-1)}{2}x^2-(1-x)^n.$$ Indeed, $$f'(x)=-n+n(n-1)x+n(1-x)^{n-1}$$ and $$f''(x)=n(n-1)-n(n-1)(1-x)^{n-2}=n(n-1)\left(1-(1-x)^{n-2}\right)>0.$$ Thus, $$f'(x)>f'(0)=0,$$ $$f(x)>f(0)=0$$ and the lemma is proven.

Now, take $$x=\frac{1}{100^2}$$ and $$n=100.$$

• What about $0.9999^{101}<0.99?$ May 28 '19 at 4:30
• @J. W. Tanner I added something. See now. May 28 '19 at 4:33
• Thank you very much May 28 '19 at 4:34
• How to show that $\left(1-\frac{1}{100^2}\right)^{100}<1-\binom{100}{1}\cdot\frac{1}{100^2}+\binom{100}{2}\cdot\frac{1}{100^4}$. May 28 '19 at 4:35
• @zongxiang yi I added something. See now. May 28 '19 at 4:46

Here is an outline of a proof that $$(1-x)^{\left(1-\frac{1}{x}\right)}<(1+x)^{\left(\frac{1}{x}\right)}$$ when $$0<|x|<1,$$

using some of your tags such as derivatives and logarithms.

Let $$f(x)=\dfrac{(1+x)^{\frac1x}}{(1-x)^{(1-\frac1x)}}=\dfrac{(1-x^2)^{\frac1x}}{1-x}.$$

Note that $$f(0)$$ is not defined, but $$\lim_{x\to0}f(x)=1$$.

$$f'(x)=-(1-x^2)^{\left(\frac1x-1\right)}\left[1+\dfrac{1+x}{x^2}\ln(1-x^2)\right].$$

For $$0, $$\ln(1-x^2)<-x^2<-\dfrac{x^2}{1+x}$$, so $$f'(x)>0.$$

For $$-1-\dfrac{x^2}{1+x},$$ so $$f'(x)<0.$$

It follows that $$f(x)>1$$ when $$0<|x|<1,$$

and thus $$(1-x)^{\left(1-\frac{1}{x}\right)}<(1+x)^{\left(\frac{1}{x}\right)}$$ when $$0<|x|<1.$$