# Summation x(1/2)^x

1. Mar 8, 2008

### kuahji

I have the summation x(1/2)^x for (x=1,2,3,4,...)

So I set it up as s=1(1/2)+2(1/4)+3(1/8)+4(1/16)...

This is however where I'm lost, I'm not exactly sure how to sum an infinite sequence, it hasn't really been introduced in any of my math courses, it just popped up in a statistics problem though.

The book shows the next step as 1/2s=1/4+2(1/8)+3(1/16)...
then
1/2s=1/2+1/4+1/8+1/6+...=1
Thus s=2

But I'm not sure exactly what they are doing in the first & second step there. Any explanations would be helpful.

2. Mar 8, 2008

### Dick

In the second step they just multiplied s by (1/2). In the next step they subtract the second series for (1/2)s from the first series for s. E.g. the 1/8 in the last series comes from 3(1/8) in the first series minus 2(1/8) in the second series. FInally, the summed the geometric series in powers of 1/2.

3. Mar 8, 2008

### HallsofIvy

In the first step, assuming you mean (1/2)s and not 1/(2s), they just multiplied everything by 1/2, obviously: (1/2)s= (1/2)(1/2)+(1/2)(2/4)+ (1/2)(3/8)+ (1/2)(4/16)+ (1/2)(5/32)+ (1/2)(6/64)+ ... Now, how they got the next equation is beyond me! I thought at first that they were just adding "pairs" of fractions: 1/4+ 2(1/8)= 1/4+ 1/4= 1/2 obviously, but then 3/16+ 4/32= 3/16+ 2/32= 5/32, NOT 1/4! Perhaps they are adding additional terms but I can't find them.

The result, x= 2 is, however, correct. Here's how I would do it:

(If you don't mind I am going to use "n" for your "x" so I can use x as a variable.)

I notice that the derivative of xn is n xn-1 so the derivative of
(1/2)xn, evaluated at x= 1/2, is (1/2)n (1/2)n-1= n(1/2)n.

The point of that is that $d[(1/2) \sum x^n]/dx]$ is $\sum n x^{n-1}$ which, at x= 1/2 is the sum you want.

That helps because $\sum x^n$ is a geometric series and its sum is 1/(1-x) so $(1/2) \sum x^n= 1/(2(1-x))$. The derivative of (1/2)(1-x)-1 is (1/2)(1-x)-2 and so the original sum is (1/2)(1- 1/2)2= 2.

Last edited by a moderator: Mar 8, 2008