Calc6.24

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Discuss the convergence or divergence of the series \sum _{{n=1}}^{{\infty }}\sin {\frac  {1}{n}}.

Let’s use the limit comparison test.

\lim _{{n\rightarrow \infty }}{\frac  {\sin {\frac  {1}{n}}}{{\frac  {1}{n}}}}=\lim _{{k\rightarrow 0}}{\frac  {\sin k}{k}}=1\,

This limit is known to be 1. If you do not see this, use L’Hospital’s Rule. Thus, by the limit comparison test, since \sum _{{n=1}}^{{\infty }}{\frac  {1}{n}} diverges, then our series also diverges.


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