Find the greater number: $1000^{1000}$ or $1001^{999}$

My Attempt:

I know that: $(a+b)^n \geq a^n + a^{n-1}bn$.

Thus, $(1+999)^{1000} \geq 999001$

And $(1+1000)^{999} \geq 999001$

But that doesn't make much sense.

I want some hints regarding how to solve this problem.



Look at the quotient $$ \frac{1001^{999}}{1000^{1000}}=\frac1{1001}\underbrace{\left(1+\frac1{1000}\right)^{1000}}_{\approx e}$$

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    $\begingroup$ Wow, striking. +1 $\endgroup$ – Alex Wertheim May 14 '14 at 6:25
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    $\begingroup$ @Hagen The approach looks nice. But how do I use it to get the answer? I am a bit newbie so please help me out :) $\endgroup$ – Gaurang Tandon May 14 '14 at 6:28
  • $\begingroup$ This is the ratio of the numbers you are comparing. The top is apparently much smaller than the bottom, around $e/1001$ times smaller. $\endgroup$ – orion May 14 '14 at 7:53
  • $\begingroup$ @GaurangTandon This pretty much is the answer... the thing on the right is less than one. Let's call it $c$. Then $1001^{999} = c1000^{1000}$, where $c$ is around $e/1001$. $\endgroup$ – user98602 May 14 '14 at 7:53
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    $\begingroup$ @GaurangTandon The approximation is certainly $<1001$, which is all we need here. More rigorously (but awfully wasteful), $(1-\frac1{1001})^{1000}>1-\frac{1000}{1001}$ by Bernoulli's inequality, hence the reciprocal $(1+\frac1{1000})^{1000}$ is $<1001$ $\endgroup$ – Hagen von Eitzen Jan 25 '16 at 13:32

$\forall r \in \Bbb N-\{1\}$, we have by applying the AM-GM inequality to the $r$ numbers $r-1$ of which equal $r+1$ and one $1$, we have, $$\frac {1+(r-1)(r+1)}{r} \gt (1 \times (r+1)^{r-1})^\frac {1}{r}$$ wherefrom we have, $r^r \gt (r+1)^{r-1}.$

  • $\begingroup$ What is the AM-GM inequality ? $\endgroup$ – Gaurang Tandon May 14 '14 at 6:41
  • $\begingroup$ See this page, it is a very famous inequality... en.wikipedia.org/wiki/… $\endgroup$ – Indrayudh Roy May 14 '14 at 6:44
  • $\begingroup$ 2 questions: 1) What is the symbol ? 2) How did you actually use the AM-GM inequality ? $\endgroup$ – Gaurang Tandon May 14 '14 at 6:50
  • $\begingroup$ $ \forall$ mean for all, i.e. for every element $r\in \Bbb N-\{1\}$. $\endgroup$ – Indrayudh Roy May 14 '14 at 6:52

Divide the two numbers and prove the result is grater or less than 1. Then, respectively, the first number or the second number is bigger than the other one.

This is my first post here, so I hope somebody will help me with the proper formatting, but generally its like this:

$$ \frac{1000^{1000}}{1001^{999}} = \frac{1}{1001}\cdot \frac{1000^{1000}}{1001^{1000}} = \frac{1}{1001}\cdot \left( \frac{1000}{1001}\right)^{1000} $$ As you can see here: $\frac{1}{1001} < 1$ and $(1000/1001)^{1000} < 1$, so $1000^{1000}/1001^{999} < 1$ So from here we can conclude that

$1000^{1000} < 1001^{999}$

  • $\begingroup$ I formatted the equations for you, which makes it easier to see that you've made an algebraic mistake when you pull out the $1/1001$ factor. $\endgroup$ – mrf May 14 '14 at 6:52
  • $\begingroup$ Hey, thank you, now I see that i made a mistake :D I'll try to correct it, but until then its wrong. Sorry, maybe I need a cup of coffee :D ;) $\endgroup$ – LGalabov May 14 '14 at 6:55
  • $\begingroup$ You can delete your post while it's wrong, then undelete it once you've fixed it. That way you avoid confusing people and you also avoid downvotes due to the mistake. $\endgroup$ – MvG May 14 '14 at 7:18
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    $\begingroup$ @MvG: Unregistered users cannot delete posts. $\endgroup$ – Asaf Karagila May 14 '14 at 8:14
  • $\begingroup$ This is wrong, the first step should multiply by $1001$ instead of $1/1001$ $\endgroup$ – Guido Aug 9 '14 at 15:01

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