Help me find this infinite summation

[island-boy]

$$\sum_{x =1}^{\infty} x (\frac{1}{2})^x$$

is there a formula for this, like for infinite geometric summation?

 

[matt grime]

Yes. Given x(1/2)^x, can you think of some other quantity that you can apply something well known to to get x(1/2)^x, or perhaps more suggestively x(1/2)^{x-1}?

 

[0rthodontist]

Explaining that a little, consider it as a special case of
$$\sum_{x =1}^{\infty} x y^x$$

If you are familiar with generating functions for sequences, multiplying every nth term by n is a standard operation.

 

[island-boy]

I think, I get what you are trying to say...

since
$$\sum_{x=1}^{\infty} (\frac{1}{2})^x = 1$$ by the infinite geometric summation formula

then, is it correct that:
$$\sum_{x=1}^{\infty} x(\frac{1}{2})^x = x$$ ?

 

[StatusX]

No. Try replacing 1/2 with a variable like y as Orthodontist suggested, and differentiate the sum with respect to y.

 

[island-boy]

oh, I think I got what you are trying to say:
$$\sum_{x=1}^{\infty} y^x = \frac{y}{1-y}$$
differentiating, I get:
$$\sum_{x=1}^{\infty}x y^{x-1} = \frac{1}{(1-y)^2}$$
multiplying by y,
$$\sum_{x=1}^{\infty}x y^x = \frac{y}{(1-y)^2}$$
so,
$$\sum_{x=1}^{\infty}x (\frac{1}{2})^x = 2$$

Is this correct?

 

[0rthodontist]

Yes it is.

 

[island-boy]

great! thanks for your help, matt, orthodontist, statusX