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

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Discussion Overview

The discussion revolves around the algebraic proof of the infinite series \(\sum_{n=0}^{\infty} \frac{n}{2^n} = 2\). Participants explore various methods to derive this result, including differentiation of power series and summation techniques.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant requests help to prove the series, indicating a belief that there is a simple trick involved.
  • Another participant asks for context regarding the application of the problem, suggesting it may be a homework or self-study issue.
  • Several participants propose defining a function \(f(x)\) based on the series and taking its derivative to derive a formula for \(\sum_{n=1}^{\infty} n x^{-n}\).
  • One participant suggests using summation by parts to derive a partial sum for the series.
  • A later reply points out a potential mistake in the earlier derivation, correcting the expression for the sum of the series.
  • Another participant expresses gratitude for the responses and mentions a technical issue with the website affecting their ability to view formula renderings.

Areas of Agreement / Disagreement

Participants present multiple approaches to the problem, and while some corrections are made, there is no consensus on a single method or final answer. The discussion remains unresolved regarding the most effective proof.

Contextual Notes

Some participants express uncertainty about the definitions and steps involved in the derivations, indicating that assumptions may be missing or unclear.

morgul
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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
 
Last edited by a moderator:
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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?
 
Last edited by a moderator:
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.
 
Last edited:
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.
 
Last edited:
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}
 
Last edited:
  • #10
Yes, thank you. The website glitched on me before I could see the renderings of my formulas.
 

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