Alternating Series Convergence: Is the Series Sum of sin(1/n^2) Convergent?

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SUMMARY

The series sum of sin(1/n^2) converges, as established through the limit comparison test with the p-series 1/n^2. The limit of n^2*sin(1/n^2) approaches a finite value as n approaches infinity, confirming that the convergence of the series sum of 1/n^2 implies the convergence of the series sum of sin(1/n^2). The initial assumption of divergence was incorrect due to a misunderstanding of the series behavior at infinity.

PREREQUISITES
  • Understanding of limit comparison test in calculus
  • Familiarity with p-series convergence criteria
  • Knowledge of the behavior of the sine function as its argument approaches zero
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the Limit Comparison Test in detail
  • Review convergence criteria for p-series, specifically for p = 2
  • Explore the behavior of sin(x) as x approaches zero
  • Practice problems involving series convergence and divergence
USEFUL FOR

Students studying calculus, particularly those focusing on series convergence, as well as educators looking for examples of limit comparison tests in action.

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Homework Statement



Sorry I don't know how to use symbols on this site so bear with me:

the question is does the following series converge: sum of sin(1/n^2) where n goes from 1 to infinity

Homework Equations



Limit comparison test, maybe others

The Attempt at a Solution



Okay I think I got the solution but I'm not sure if the logic is correct. I divided sin(1/n^2) by the p-series (1/n).

This comes out to n/sin(1/n^2)...the top will go to infinity and the bottom will go to 0 as n goes to infinity, so the series divergies. Because 1/n also diverges (harmonic series) sin(1/n^2) also diverges. Is this right?
 
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Don't know why I said 'alternating' series in the thread title...my brain is becoming mush!
 
I think limit test is probably the way to go, however I think you're doing a few things wrong with it.

With a_n being the given function, consider b_n = \frac{1}{n^2}

Then,
\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} n^2\sin{\frac{1}{n^2}}

Now, you should be able to show that that limit is finite, so the convergence of the sum of b_n would imply the convergence of the sum of a_n. Does that help?
 
Yep that fixes it, I also made an algebraic error thinking sin was on the denominator. Brain is going to mush. Thanks!
 

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