Problem 9 Easy Difficulty

Determine whether the series converges or diverges.
$ \displaystyle \sum_{k = 1}^{\infty} \frac {\ln k}{k} $

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Yiming Z.

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Hello. So here we have these somewhere we have K. Going from one to infinity of the natural log of K over K. So um well we can expand this series with expect to K. So therefore we have while the natural log of 1/1 plus the natural log of 2/2 plus well and so on. Plus we get these some where we get, let's say K. Now goes from three to infinity of the natural log of K over K. So we then take the series. Here we have the K. E goes from three to infinity of Ellen K over K. And we're going to use the comparison test to determine whether the series converges or diverges. So we know that the comparison test right? If we take the some of B seven is convergent where we get that A seven is less than equal to B. Seven for all end than the sum of a servant is also convergent. And if the Saban is out of some of the seven is divergent and we get a servant is greater than or equal to B. Seven for all N. Than the sum of a servant is also divergent. So here, since we have that Ellen K is greater than one for all K, greater than equal to three, it follows that we have the natural log of K over K is going to be greater than one over K for all K greater than or equal to three. So here let's let um a sub K equals the natural log of K over K. And let's let be sub K equal are one over K. Now, since each term of the series um some going from K equals three to infinity of L. N. F. K over K is greater than the corresponding terms of the series where K goes from three to infinity of one over K. Which we have as a P series with P equal to one, which is divergent. Therefore, we had at the series K, going from three to infinity of Ellen of K over K must be divergent as well, so must be divergent by the comparison test.