Does (lnk)/(k^3), where as k starts at one and continues on to infinity, converge? ...?

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asked Mar 7, 2017 4.8k views

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Russel · Answer 1 · 2017-03-13T22:11:14+0000

Yes. The numerator does not increase as fast as the denominator increases, causing the function's value to decrease with every subsequent increase in the value of k. This causes the function to converge at a point.

Does (lnk)/(k^3), where as k starts at one and continues on to infinity, converge? ...?

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