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

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 $$0. I used $$\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!

• Please don't write questions as mere problem statements. Your first version was completely fine as you showed what you tried (something this site highly values!). Therefore I rollbacked your question to the first version. Mar 19, 2020 at 4:54
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## 2 Answers

HINT:

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

• @TheStudent For instance, yes. Note, however, that you do not need the first inequality as the terms are strictly positive. The only thing you need is to bound it from above by a convergent series. Think about when $\frac{\ln(n)}{e^n}<\frac{n}{e^n}$ holds instead. Mar 19, 2020 at 4:32

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 $$n-1 <2^n$$.