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# Determine whether the series converges or diverges.$\displaystyle \sum_{n = 1}^{\infty} \frac {e^{1/n}}{n}$

## $\sum_{n=1}^{\infty} \frac{e^{1 / n}}{n}$ diverges by the direct comparison test

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##### Kristen K.

University of Michigan - Ann Arbor

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let's determine whether the Siri's converges of averages well, either the one over end since one over end is always bigger than zero. Eat to the one of her end is bigger than E to the zero, which equals one. So here I can take our Siri's and say, This is not that way. I should do other inequality. This is bigger than or equal to the sum from one to infinity of just one over end sense E to the one over end. It's bigger than one, so we're just replacing the numerator eat to the one over and was something smaller, So the fraction is a hole gets smaller. And how about the Siri's here of one over end the Siri's diverges. This is known as the harmonic series, and the reason and diverges were Book proves it. But another way to prove it is just to use the pee test. And here P equals one that's the power of and in the denominator. And any time this number is less than or equal to one, you'LL will have divergence. So that's not for our Siri's. This is for the Green Series, the one over and squared Hope or the excuse me the one over and that someone there now to explain why our Siri's diverges since either the one over and over and his positive these are always bigger than zero. We've just shown that this Siri's the one in question, each of the one over and over and diverges bye, the computers and test. We just compared our Siri's with the smaller Siri's the lower bound, which would happen to be harmonic series that divers, so by comparison, are larger. Siri's and Red also has toe diverge, and that's your final answer.

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