Convergence Tests: Test for Convergence of Series

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Homework Help Overview

The discussion revolves around testing the convergence of the series \(\sum \frac{\sin(1/n)}{\sqrt{\ln(n)}}\) from \(n=2\) to infinity, focusing on the application of the limit comparison test and integral test.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the comparison of \(\sin(1/n)\) with \(1/n\) and \(1/(2n)\) to assess convergence. Questions arise regarding the validity of using these comparisons and the implications of the integral test.

Discussion Status

Participants are actively engaging with different comparison series and testing their implications for convergence. There is a recognition of the need for a series that is less than the original to support divergence claims, and some participants express skepticism about external computational tools like Wolfram Alpha.

Contextual Notes

There is an ongoing debate about the accuracy of computational results from Wolfram Alpha, with participants questioning its reliability in the context of convergence tests.

craig16
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Homework Statement



Test for convergence

[tex]\sum[/tex] sin(1/n) / [tex]\sqrt{ln(n)}[/tex]

sum from 2 to inf

Homework Equations



Limit comparison test

if lim an / bn to inf
doesn't equal 1 you know if it converges or diverges
by limit comparison test

The Attempt at a Solution



I've tried a lot of different things for bn, I think the only way to test this series for convergence is from the limit comparison test.

I know the series converges I just can't prove it
 
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How does sin(1/n) compare with 1/n? How about with 1/(2n)? Think integral test with comparison series like that.
 
Last edited:
So , 1/n is always greater than sin(1/n)

If I just replace sin(1/n) with 1/n , I have a "an" that is less than the original and the new function diverges by the integral test.

So the sum is divergent by the comparison test?
 
craig16 said:
So , 1/n is always greater than sin(1/n)

If I just replace sin(1/n) with 1/n , I have a "an" that is less than the original and the new function diverges by the integral test.

So the sum is divergent by the comparison test?

You've got a divergent series that's greater than the original. That doesn't prove divergence. You need one that's less than the original. That's why I also suggested 1/(2n).
 
Yes, Wolfram Alpha is quite wrong. Unless you believe 962 is the same as infinity.
 
Awesome :)

Thank you, I've spent way too much time thinking about that sum, lol.
 
craig16 said:
Okay, I see how the test would work replacing sin(1/n) with 1/2n making the new equation always less than the original. Then the integral test works and the sum is divergent.

Does that mean that wolfram is wrong though?

http://www.wolframalpha.com/input/?i=sum from 2 to inf (sin(1/n))/(sqrt(ln(n)))&t=macw01

When I plug the sum in it tells me it converges to 962 or so.

A good way to test if Wolfram-Alpha is lying is to click the More Digits button. If the number changes when you do that, W-A is probably lying to you.
 

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