This question is ruining my head?

Question

If r<1 and positive, and m is a positive integer, show that
(2m + 1)r^m( 1-r ) < 1-r^2m+1.

Answer 1

The clarification to the answer above is critical. If the second interpretation is correct, then this is not always true - try subsituting 0.99 for r and 1 for m.

I assumed on the LHS this is r^[m(1-r)] and on the RHS this is r^(2m+1).

Please clarify if this is incorrect.

Answer 2

What do you mean by r^m( 1-r ) on the LHS? Is this r^m times (1-r), or r^[m(1-r)] ???

Same question for RHS: is this r^2 * (m+1), r^(2m+1), r^(2m) + 1, or what?

Answer 3

(2m + 1)r^m( 1-r ) < 1-r^2m+1

Let m=2 and r =.5
Then 2m+1 = 2*2+1=5
r^m(1-r) = .5^2(1-.5)=.5
1-r^(2m+1) = 1- .5^5= .96875
So 5(.5)<.96875
2.5 < .96875
This is false so we have shown by counterexample that the inequality does not hold.

Answer 4

I think what you want to do is make r a constant and take the limit of both as m increases from 1 to 2, to 3, infinitely. Then go back and make m a constant, and change r around, and work another limit.

Answer 5

Divide both sides by (1-r). The RHS becomes 1+r+.......+r^2m.
Now for each k
2 r^m < r^(m-k) + r^(m+k),

because the geomatric mean is always less than the arithmetic mean. You sum and that's it.

Answer 6

what the ^%&£