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.

Related questions
See all questions in Limit Comparison Test for Convergence of an Infinite Series
Impact of this question
24346 views around the world
You can reuse this answer
Creative Commons License