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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!

mrtaurho
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user758896
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2 Answers2

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HINT:

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

mrtaurho
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    @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. – mrtaurho Mar 19 '20 at 04:32
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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 $n-1 <2^n$.

N. S.
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