How to choose the Bn for limit comparison test?

If the An is #(e^(1/n))/n#
how would you determine what bn to use to compare with this?

1 Answer
Bill K.
Dec 22, 2017

Note that #e^{1/n}>1# for all integers #n>0#. Therefore, we expect that #sum_{n=1}^{infty}e^{1/n}/n# will diverge. Try comparing it to the divergent harmonic series #sum_{n=1}^{infty}1/n# to show this with the limit comparison test (so use #b_{n}=1/n#).

Explanation:

Let #a_{n}=e^{1/n}/n# and #b_{n}=1/n#, noting that #a_{n} > b_{n} > 0# for all integers #n>0#.

Now compute #lim_{n->infty}a_{n}/b_{n}#. We are"hoping" it is a positive number and not #infty#, which will allow us to say that #sum_{n=1}^{infty}e^{1/n}/n# diverges by the Limit Comparison Test since we know that the harmonic series #sum_{n=1}^{infty}1/n# diverges.

But clearly, #lim_{n->infty}a_{n}/b_{n}=lim_{n->infty}e^{1/n}=1#, a positive number (and not #infty#). We are done.

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