How to demonstrate sigma(i/2^i)<=2 (i=1 to n)

问题

How to demonstrate following Unequal:

sigma(i/2^i)<=2 (i=1 to n)

回答1:

If we look at the series then it looks like:
We consider n = infinity for maximum value of sum.
S = 1/2 + 2/4 + 3/8 + 4/16 + 5/32 +.... + 0 -(1)
Clearly,
S/2 = 1/4 + 2/8 + 3/16 + 4/32 + ---- + 0 -(2)

Subtracting (1) from (2) we get:
S/2 = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 +... + 0
In this a = 1/2 and r = 1/2 so sum of this infinite gp is a/(1-r) = 1/2/(1-1/2) = 1

So maximum value of S/2 is 1
then maximum value of S is 2 or S <= 2

回答2:

The sum of a + ar + ar^2 + ar^3 + ... is given by a / (1 - r). If only a finite number of terms are present, there is a non-negative remainder, that is, the sum will be [ a / (1 - r) ] - R. Plugging in the values of a and r, i get 2 - R. It seems that this is consistently <= 2. Q.E.D.

来源：https://stackoverflow.com/questions/38983113/how-to-demonstrate-sigmai-2i-2-i-1-to-n