Look at the following integral
That iintegral has as values
I(2) = -2пUP>2 + 24
I(3) = -24пUP>2 + 240
I(4) = 2пUP>4 - 360пUP>2 + 3360
I(5) = 60пUP>4 - 6720пUP>2 + 60480
I(6) = -2пUP>6 + 1680пUP>4 - 151200пUP>2 + 1330560
All sums of powers of п. We will determine a maximum for the integral:
x(п-x) ≤ 0.5п(п-0.5п
0.5п(п-0.5п
= 0.25пUP>2 0.25пUP>2 ≤ 3
x(п-x) ≤ 3
Now the other part:
0 ≤ sin(x) ≤ 1
sin(x) ≤ 1
Filling this in and calculating the integral gives you an upperlimit for I(n) of п*3UP>n/n!
Now let's assume п is rational (and repeats): п=p/q, where p and q are two whole numbers and the expression has been simplified as much as possible.
That means for example with I(4):
(2pUP>4 - 360pUP>2qUP>2 + 3360qUP>4)/qUP>4
Because this is an positive numbers that has been simplified as much as possible we can say:
1/qUP>4 ≤ (2pUP>4 - 360pUP>2qUP>2 + 3360qUP>4)/qUP>4
This is true for every I(n), so you can say:
1/qUP>n ≤ I(n)
We determined a upper limit for I(n), so 1/qUP>n is always smaller than that:
1/qUP>n ≤ п*3UP>n/n!
1 ≤ п*(3q)UP>n/n!
But lim(x->∞
xUP>n/n! goes to 0. That would mean: lim(x->∞
1 ≤ п*(3q)UP>n/n! gives
1 ≤ 0
This is ofcourse not true. That problem is our assumption that п is rational. Q.E.D.
[Edited on 9-4-2004 by amantine]


