Finding the Sum of a Power Series

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Homework Help Overview

The original poster attempts to find the sum of a power series involving alternating terms and a variable raised to the power of n. The series in question is expressed as the sum of (-1)^n nx^n from n=0 to infinity.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between the original series and known power series expansions, particularly focusing on how to incorporate the (-1)^n factor into the existing series manipulations.

Discussion Status

Some participants have provided insights into relevant series expansions, particularly the geometric series and its variations. There is an ongoing exploration of how these expansions relate to the original problem, with no explicit consensus reached yet.

Contextual Notes

The discussion includes attempts to manipulate series and clarify the application of alternating series, but it does not resolve the original poster's query regarding the incorporation of the (-1)^n term.

student45
I'm trying to find the sum of this:

[tex] \[<br /> \sum\limits_{n = 0}^\infty {( - 1)^n nx^n } <br /> \][/tex]

This is what I have so far:

[tex] \[<br /> \begin{array}{l}<br /> \frac{1}{{1 - x}} = \sum\limits_{n = 0}^\infty {x^n } \\ <br /> \frac{1}{{(1 - x)^2 }} = \sum\limits_{n = 0}^\infty {nx^{n - 1} } = \sum\limits_{n = 1}^\infty {nx^{n - 1} } \\ <br /> \frac{x}{{(1 - x)^2 }} = \sum\limits_{n = 1}^\infty {nx^n } \\ <br /> \end{array}<br /> \][/tex]

So how do I get the (-1)^n part in there? Any suggestions would be really helpful. Thanks.
 
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[tex]\frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^{n}x^{n}[/tex]
 
Last edited:
Oh, I see.. Where exactly does that come from?
 
[tex]\frac{1}{1+x} = \frac{1}{1-(-x)} = \sum_{n=0}^{\infty} (-x)^{n} = (-1)^{n}x^{n}[/tex]
 
Ah! Of course. Okay. Thanks.
 

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