Question:
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.
Thanks.
$\endgroup$ 53 Answers
$\begingroup$Look at the quotient $$ \frac{1001^{999}}{1000^{1000}}=\frac1{1001}\underbrace{\left(1+\frac1{1000}\right)^{1000}}_{\approx e}$$
$\endgroup$ 8 $\begingroup$$\forall r \in \Bbb N-\{1\}$, we have by applying the AM-GM inequality to the $r$ numbers $r-1$ of which equal $r+1$ and one $1$, we have, $$\frac {1+(r-1)(r+1)}{r} \gt (1 \times (r+1)^{r-1})^\frac {1}{r}$$ wherefrom we have, $r^r \gt (r+1)^{r-1}.$
$\endgroup$ 4 $\begingroup$Divide the two numbers and prove the result is grater or less than 1. Then, respectively, the first number or the second number is bigger than the other one.
This is my first post here, so I hope somebody will help me with the proper formatting, but generally its like this:
$$ \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
$1000^{1000} < 1001^{999}$
$\endgroup$ 5
