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Let $a = (10^n - 1)^{(10^n)}$ and $b=(10^n)^{(10^n - 1)}$

Which of these numbers is greater as n gets large?

I believe it is $a$ after looking at some smaller special cases, but I'm not sure how to prove it.

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4 Answers 4

up vote 37 down vote accepted


$$\frac{a}{b} = \frac{(10^n-1)^{10^n}}{(10^n)^{10^n-1}} = (10^n-1)\left(1- \frac{1}{10^{n}}\right)^{10^n-1} \approx (10^n-1)e^{-1} > 1.$$

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Because the logarithm is monotonic, you can take the difference of the logs: $$\log a - \log b=10^n \log(10^n-1)-(10^n-1)\log 10^n\\ =10^n(\log(10^n)+\log(1-\frac 1{10^n})-\left(n(10^n-1)\log 10\right) \\ \approx n10^n\log(10)-1-n10^n\log(10)+n\log(10)\\ =n\log(10)-1 \gt 0$$

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Say $x={ 10 }^{ n }\\$ $$\frac { { \left( x-1 \right) }^{ x } }{ { x }^{ x-1 } } =x{ \left( \frac { x-1 }{ x } \right) }^{ x }=x{ \left( 1-\frac { 1 }{ x } \right) }^{ x }\\ \lim _{ x\rightarrow \infty }{ x{ \left( 1-\frac { 1 }{ x } \right) }^{ x } } =\lim _{ x\rightarrow \infty }{ \frac { x }{ e } } =\infty $$

According to calculation on Wolframalpha for any $x > 3.2932$, $$(x-1)^{ x }>x^{x-1}$$

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Let $x=10^n$, $y=10^n-1$. The function $\frac {x} {\ln x}$ is increasing from $e$.

Both $x$ and $y$ are greater than $e$, so $\frac {x} {\ln x} > \frac {y} {\ln y} \Rightarrow x\ln y > y\ln x \Rightarrow y^x > x^y$

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