Question

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`).

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.