Series Convergence/Divergence Proofs

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SUMMARY

The series \(\sum_{n=1}^{\infty} n \sin(\frac{1}{n})\) diverges as proven by applying the Nth term test and L'Hôpital's rule. The series was rewritten as \(\sum_{n=1}^{\infty} \frac{\sin(\frac{1}{n})}{\frac{1}{n}}\). By evaluating the limit \(\lim_{n\to\infty} \frac{\cos(\frac{1}{n})\frac{-1}{n^2}}{\frac{-1}{n^2}}\), the terms \(\frac{-1}{n^2}\) cancel, leading to \(\lim_{n\to\infty} \cos(\frac{1}{n}) = 1\). This confirms that the series diverges according to the Nth term test.

PREREQUISITES
  • Understanding of series convergence and divergence
  • Familiarity with L'Hôpital's rule
  • Knowledge of the Nth term test for series
  • Basic calculus concepts, particularly limits
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  • Study advanced techniques for proving series convergence and divergence
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  • Explore the implications of the Nth term test in various series
  • Investigate other convergence tests such as the Ratio Test and Root Test
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chimychang
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\sum_{n=1}^{\infty} n \sin(\frac{1}{n})

I rewrote the sum as \sum_{n=1}^{\infty} \frac{\sin(\frac{1}{n})}{\frac{1}{n}}

Then I applied the Nth term test and used L'Hoptials rule so \lim_{n\to\infty} \frac{\cos(\frac{1}{n})\frac{-1}{n^2}}{\frac{-1}{n^2}}

The \frac{-1}{n^2} cancel out and the lim_{n\to\infty} \cos(\frac{1}{n}) is 1 which by the nth term test is divergent. Is that a legitimate proof of divergence?
 
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