Home > Calculus > Limit Comparison Test for Convergence of an Infinite Series

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?

Answer 1

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#).

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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Answer 2

Charles Anctil

To choose $B_n$ for the limit comparison test, follow these steps:

Select a series $\sum a_n$ that you suspect behaves similarly to the series $\sum b_n$ .
Choose $B_n$ such that $0 < B_n < a_n$ for all $n$ beyond some point.
Compute the limit $\lim_{n \to \infty} \frac{a_n}{B_n}$ .
If the limit is a positive finite number, then $\sum a_n$ and $\sum B_n$ either both converge or both diverge.
If the limit is zero or infinity, consider choosing a different $B_n$ and repeat the process until the limit meets the conditions outlined in step 4.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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?

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