Proving \sum{n=0=>inf} (n / 2^n) = 2

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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
 
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morgul said:
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

Welcome to the PF. What is the context of your question? What is the application?
 
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berkeman said:
Welcome to the PF. What is the context of your question? What is the application?
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.
 
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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.
 
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.
 
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.
 
Alternatively, summation by parts yields partial sum
<br /> \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})=...<br />
 
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Alternatively, summation by parts yields partial sum
<br /> \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})=...<br />
Sorry for double post.
 
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hamster143 said:
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&#039;(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.

You've made a slight mistake.

\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}
 
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Yes, thank you. The website glitched on me before I could see the renderings of my formulas.
 
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