# Series proof

1. Jun 26, 2010

### morgul

Hey,

I am wondering if anyone could algebraically prove that $$\sum{n=0=>inf} (n / 2^n)$$ = 2. There is some simple trick to it but I am stumped. :/

Here is a clearer picture of my summation:
http://img638.imageshack.us/img638/2114/texer1.png [Broken]

Last edited by a moderator: May 4, 2017
2. Jun 26, 2010

### Staff: Mentor

Welcome to the PF. What is the context of your question? What is the application?

Last edited by a moderator: May 4, 2017
3. Jun 27, 2010

### Redbelly98

Staff Emeritus
I, too, wish to know this. morgul, could you respond?

If this is a homework problem, or even a self-study problem from a book, we have rules about getting help on a problem like this. Just click the "Rules" link at the top of this page, and then scroll down to the section titled Homework Help at the rules page.

Last edited: Jun 27, 2010
4. Jun 28, 2010

### morgul

Ah, thank you for the replies! This is not a homework problem. This was from an old notebook of mine and the solution is incomplete. It was just bothering me.

5. Jun 29, 2010

### hamster143

Define

$$f(x) = \sum_{n=0}^{\infty} x^{-n} = 1 + 1/x + 1/{x^2} + ... = 1/(x-1).$$

Take a derivative:

$$- 1/(x-1)^2 = f'(x) = - \sum_{n=0}^{\infty} n x^{-n-1} = - (1/x) \sum_{n=1}^{\infty} n x^{-n}.$$

$$\sum_{n=1}^{\infty} n x^{-n} = x/(x-1)^2$$

Set x=2.

6. Jun 29, 2010

### hamster143

Define

$$f(x) = \sum_{n=0}^{\infty} x^{-n} = 1 + 1/x + 1/{x^2} + ... = 1/(x-1).$$

Take a derivative:

$$- 1/(x-1)^2 = f'(x) = - \sum_{n=0}^{\infty} n x^{-n-1} = - (1/x) \sum_{n=1}^{\infty} n x^{-n}.$$

$$\sum_{n=1}^{\infty} n x^{-n} = x/(x-1)^2$$

Set x=2.

7. Jun 29, 2010

### losiu99

Alternatively, summation by parts yields partial sum
$$\sum_{n=1}^N \frac{n}{2^n}= -\sum_{n=1}^{N-1} 2\left[(\frac{1}{2})^n-1\right]+2N(1-2^{-N})=...$$

Last edited: Jun 29, 2010
8. Jun 29, 2010

### losiu99

Alternatively, summation by parts yields partial sum
$$\sum_{n=1}^N \frac{n}{2^n}= \sum_{n=1}^{N-1} 2\left[(\frac{1}{2})^n-1\right]+2N(1-2^{-N})=...$$
Sorry for double post.

Last edited: Jun 29, 2010
9. Jun 29, 2010

### Mute

$$\sum_{n=0}^\infty x^{-n} = \frac{1}{1-\frac{1}{x}} = \frac{x}{x-1}$$

So,

$$\sum_{n=0}^\infty \frac{n}{x^n} = -x\frac{d}{dx}\left(\frac{x}{x-1}\right) = \frac{x^2}{(x-1)^2} - \frac{x}{x-1}$$

Last edited: Jun 29, 2010
10. Jun 29, 2010

### hamster143

Yes, thank you. The web site glitched on me before I could see the renderings of my formulas.