Use the Comparison Test or Limit Comparison Test to determine the convergence of $\sum_{n=1}^ \infty \frac{\ln(n)}{e^n}$

Question

I know using the ratio test that this series does in fact converge, but I am stuck with how to move through the problem as it requires the use of one of the comparison tests.
I used the limit comparison test, where $\lim_{x \to \infty} \frac{a_n}{b_n} = L$ where $0<L<\infty$. I used $\frac{\ln(n)}{e^n}$ as my $a_n$ and $\frac{1}{e^n}$ as my $b_n$. This equals $\lim\limits_{n \to \infty} \frac{\ln(n)}{e^n}\frac{e^n}{1}$ which is the same as $\lim\limits_{n \to \infty}\ln(n)$. But this goes to infinity, meaning the limit comparison test cant work.
The limit of $\ln(n)$ is not a finite number as it diverges. When trying the regular comparison test, I usually find that is $\sum a_n$ is greater than $\sum b_n$ for all $n > 2.7$. But $b_n$ is a convergent geometric series which does not fulfill the conditions of the regular comparison test.

Any help is appreciated!

Answer 1

HINT:

We have $\lim\limits_{n\to\infty}\frac{\ln(n)}n=0$. How can you use this to construct a convergent majorant?

Answer 2

Hint

It is easy to see that for $x>0$ we have $$\ln(x+1)< x$$ see for example here

This gives $$\frac{\ln(n)}{e^n} < \frac{n-1}{e^n}$$

Now, it is an easy exercise to argue that $n-1 <2^n$.