如何演示 sigma(i/2^i)<=2 (i=1 to n)

如果我们看一下系列，它看起来像:
我们认为总和的最大值为 n = 无穷大.
S = 1/2 + 2/4 + 3/8 + 4/16 + 5/32 +.... + 0 - (1)
很明显，
S/2 = 1/4 + 2/8 + 3/16 + 4/32 + ---- + 0 - (2)

从(1)中减去(2)我们得到:
S/2 = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 +... + 0
在这个 a = 1/2 和 r = 1/2 所以这个无限 gp 的总和是 a/(1-r) = 1/2/(1-1/2) = 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 (2) from (1) 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

所以S/2的最大值是1
那么S的最大值是2或S

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