# How to choose the Bn for limit comparison test?

## If the An is $\frac{{e}^{\frac{1}{n}}}{n}$ how would you determine what bn to use to compare with this?

Note that ${e}^{\frac{1}{n}} > 1$ for all integers $n > 0$. Therefore, we expect that ${\sum}_{n = 1}^{\infty} {e}^{\frac{1}{n}} / n$ will diverge. Try comparing it to the divergent harmonic series ${\sum}_{n = 1}^{\infty} \frac{1}{n}$ to show this with the limit comparison test (so use ${b}_{n} = \frac{1}{n}$).
Let ${a}_{n} = {e}^{\frac{1}{n}} / n$ and ${b}_{n} = \frac{1}{n}$, noting that ${a}_{n} > {b}_{n} > 0$ for all integers $n > 0$.
Now compute ${\lim}_{n \to \infty} {a}_{n} / {b}_{n}$. We are"hoping" it is a positive number and not $\infty$, which will allow us to say that ${\sum}_{n = 1}^{\infty} {e}^{\frac{1}{n}} / n$ diverges by the Limit Comparison Test since we know that the harmonic series ${\sum}_{n = 1}^{\infty} \frac{1}{n}$ diverges.
But clearly, ${\lim}_{n \to \infty} {a}_{n} / {b}_{n} = {\lim}_{n \to \infty} {e}^{\frac{1}{n}} = 1$, a positive number (and not $\infty$). We are done.