Infinite sum converge to what value?

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Homework Help Overview

The discussion revolves around the convergence of the infinite series (-1)^n(x/n) from n=1 and seeks to determine its specific value. The context involves logarithmic functions and alternating series.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the series and the natural logarithm function, particularly focusing on the series representation of ln(1+x). Questions arise regarding the substitution of x and the implications for convergence.

Discussion Status

The discussion includes attempts to clarify the value of the series and its convergence properties. Some participants offer insights into the manipulation of series terms, while others question the correctness of these manipulations, leading to a productive exchange of ideas.

Contextual Notes

There is mention of potential errors in the series manipulation due to differing starting indices for the series, which may affect the interpretation of convergence and value.

pivoxa15
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Homework Statement


The infinite series (-1)^n(x/n) from n=1 converges. But what is the specific value of it?
 
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[tex]\ln(1+x) = \sum^{\infty}_{n=0} \frac{(-1)^n}{n+1} x^{n+1}[/tex]
[tex]\sum_{n=1}^{\infty} (-1)^n\frac{x}{n} = x\sum_{n=1}^{\infty} \frac{(-1)^n}{n}=x\log_e 2[/tex]
 
Gib Z said:
[tex]\ln(1+x) = \sum^{\infty}_{n=0} \frac{(-1)^n}{n+1} x^{n+1}[/tex]
[tex]\sum_{n=1}^{\infty} (-1)^n\frac{x}{n} = x\sum_{n=1}^{\infty} \frac{(-1)^n}{n}=x\log_e 2[/tex]

You have put x=1 so it should just be ln(2)?
 
Yup. Exactly.
 
But ln(2)>0 and [tex]\sum_{n=1}^{\infty} (-1)^n\frac{x}{n}<0[/tex] since the first term is negative and has the largest magnitude so will dominate the series. The series should equal -ln(2) so you may have made an error with your series manipulation.
 
Last edited:
Yea sorry about that >.< I made a mistake with the starts of the series, some were n=1 and others n=0, and I didn't handle them well. But youve got the idea
 

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