Does the series \sum_{n=1}^\infty sin(\frac{1}{n^2}) converge?

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SUMMARY

The series \(\sum_{n=1}^\infty \sin\left(\frac{1}{n^2}\right)\) converges. This conclusion is reached by applying the comparison test, specifically comparing it to the convergent series \(\sum_{n=1}^\infty \frac{1}{n^2}\). Since \(|\sin(x)| \leq x\) for \(x \geq 0\), it follows that \(\sin\left(\frac{1}{n^2}\right) \leq \frac{1}{n^2}\). Thus, the series converges due to the properties of nonnegative terms in convergence tests.

PREREQUISITES
  • Understanding of series convergence tests, particularly the comparison test.
  • Familiarity with the properties of the sine function, specifically \(|\sin(x)| \leq x\).
  • Knowledge of the p-series test, particularly the convergence of \(\sum_{n=1}^\infty \frac{1}{n^2}\).
  • Basic calculus concepts, including limits and inequalities.
NEXT STEPS
  • Study the comparison test in detail, including its applications and limitations.
  • Learn about the limit comparison test and how it can be applied to series.
  • Explore the convergence of other trigonometric series for broader understanding.
  • Investigate the behavior of \(\sin(x)\) for small values of \(x\) to understand its implications in series.
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Students studying calculus, particularly those focusing on series and convergence tests, as well as educators looking for examples of series convergence using trigonometric functions.

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Homework Statement


Does the follow serie converge:
\sum_{n=1}^\infty sin(\frac{1}{n^2})

Homework Equations


For serie a_n and b_n if:

A = 0 \leq a_n \leq b_n

if b_n converges then a_n converges

The Attempt at a Solution


I think that I have to use the equation (see 2) and then with

B = \sum_{n=1}^\infty \frac{1}{n^2}

I think that it is larger than A. However I need proof... Any suggestions.

Thanks in advance.
 
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For x>=0, |sin(x)| <= x. (Better yet, on [0,1], 0 <= sin(x) <= x.) Or you can just use the limit comparison test.

Note that both convergence tests require your series to have nonnegative terms.
 
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