Sum of a Power Series: Finding the Sum of a Series with a Variable

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Homework Help Overview

The discussion revolves around finding the sum of the power series \(\sum_{n=1}^\infty nx^{n-1}\) for \(|x|<1\). Participants explore the representation of this series and the implications of having a variable within the sum.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the known series \(\sum_{n=0}^{+\infty} x^n\) and its derivative, questioning how to derive the sum of the given series. There is an exploration of the function representation and the application of calculus techniques, such as differentiation.

Discussion Status

Some participants have offered insights into the differentiation of power series and the correct application of the chain rule. There is recognition of a mistake in the initial approach, leading to a reevaluation of the series sum.

Contextual Notes

Participants mention that a specific online platform (webassign) indicated an error in the initial function representation, prompting further discussion on the correct sum of the series.

toothpaste666
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Homework Statement



find the sum of the following series:

[itex]\sum_{n=1}^\infty nx^{n-1} , |x|<1[/itex]

Homework Equations



[itex]\frac{a}{1-r}[/itex]

The Attempt at a Solution



i know that a function representation for that series is [itex]-\frac{1}{(1-x)^2}[/itex] but how is it possible to find the sum of a series with a variable in it? please help :(
 
Last edited:
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Do you know what

[tex]\sum_{n=0}^{+\infty} x^n[/tex]

is?
 
[itex]\frac{1}{1-x}[/itex]
 
Now take derivatives.
 
i know that the series as a function is [itex]\frac{-1}{(1-x)^2}[/itex] but webassign said that was wrong. they are looking for the sum of the series.
 
toothpaste666 said:
i know that the series as a function is [itex]\frac{-1}{(1-x)^2}[/itex] but webassign said that was wrong. they are looking for the sum of the series.

Yes, it is wrong. Please show your work.
 
I wrote it as
[itex](1-x)^{-1 }[/itex]
to take the derivative i multiplied it by the exponent and subtracted one from the exponent.
[itex]-1(1-x)^{-2}[/itex]
which is
[itex]-\frac{1}{(1-x)^2}[/itex]
 
oh wait i see it now. i forgot to use the chain rule. it should be
[itex]\frac{1}{(1-x)^2}[/itex]
 
toothpaste666 said:
oh wait i see it now. i forgot to use the chain rule. it should be
[itex]\frac{1}{(1-x)^2}[/itex]

Indeed!
 
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  • #10
thanks!
 

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