Sigma [sin(1/x)] for x=1 to ∞: Converge or Diverge?

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Homework Help Overview

The discussion revolves around determining the convergence or divergence of the series Sigma [sin(1/x)] as x approaches infinity. This involves concepts from series convergence tests and the behavior of trigonometric functions as their arguments approach zero.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using the limit comparison test with the harmonic series and explore the application of L'Hospital's rule to evaluate limits. There are questions regarding the validity of results obtained through these methods.

Discussion Status

There are differing interpretations of the results from L'Hospital's rule, with some participants suggesting that the limit leads to a conclusion about divergence, while others express uncertainty about the implications of their findings. The discussion includes multiple approaches, such as the integral test, indicating a productive exploration of the topic.

Contextual Notes

Some participants question the assumptions underlying their methods, particularly regarding the behavior of sin(1/x) as x approaches infinity and the implications of comparing it to known divergent series.

Phoenix314
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I need to determine whether Sigma [sin(1/x)] for x=1 to x=infinity converges or diverges. I have a feeling that it diverges, but I don't know how to prove it.

Thanks
 
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Use the limit comparison test, with the harmonic series [itex]\sum_{n=1}^{\infty} \frac{1}{n}[/itex]
 
Thanks Cyclovenom, I took [tex]\frac{sin(1/x)}{1/x}[/tex] and did the limit as x approaches infinity with L Hopital's rule, but I got 0, so doesn't that make it inconclusive?

Thanks
 
Your result through L'Hospital is wrong, it will be 1, and sice 1 > 0, and this harmonic series diverges, then sin (1/x) diverges.

Alternatively you could consider [itex]\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1[/itex]
 
hello there

just use the integral test

[tex]\sum_{n=1}^{\infty} \sin{\frac{1}{n}} \le \int_{1}^{\infty}\sin{\frac{1}{x}}dx \not< \infty[/tex]

its not finite and so does not converge

steven
 

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