sigma of sin(1/n^3). n=1 . diverges or converges Solution The easiest solution I

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The easiest solution I am aware of is to use the limit comparison test (sometimes called asymptotic comparison test). One can compare sin (1/n) to 1/n: lim n->8 [sin (1/n^3)] / [1/n^3] If we let k = 1/n^3, then as n approaches infinity, the number k approaches 0, so we obtain lim n->8 [sin (1/n^3)] / [1/n^3] = lim k->0 [sin (k)] / k This limit is 1 (which can be seen using L'Hospital's rule, or in other ways). Since lim n->8 [sin (1/n^3)] / [1/n^3] is positive and finite, then by the limit comparison test, ? sin (1/n^3) and ? 1/n^3 either both converge or both diverge. Since ? 1/n^3 is a divergent series, then ? sin (1/n^3) is also a divergent series. Hope it will help You...

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