Limit Comparison Test for $a^n/(\sqrt{n}\cdot b^n)$

Question

Suppose I have a series like $$\sum_{n=1}^\infty \frac{5^n}{\sqrt{n}\cdot4^n}$$ and I want to use the limit comparison test to do prove if it's convergent or divergent.

I tried comparing it to the function $$f(x) = \frac{5^n}{4^n}$$ and also the function $$g(x) = \frac{1}{\sqrt{n}}$$ but none of them seem to work. I'm pretty sure I'm making being dumb right now, but I honestly can't figure out how to do this. This is a problem I made up on my own just to practice limit comparison test, and I am stumped by my own problem!

Answer 1

Doing the limit comparison test with $b_n=\frac1{\sqrt n}$ should work just fine. Taking $a_n$ as a term in your original series, you should get $\displaystyle{\lim_{n\to\infty}}\frac{a_n}{b_n}=\infty$, and since $\sum b_n$ diverges, then so does $a_n$.

That said, direct comparison is nicer and easier than limit comparison, so if you have the option to use either, why not go for the one that's less work?

Even less work, as mentioned in a comment, is simply noting that $|a_n|$ fails to approach $0$ as $n\to\infty$, which tells us immediately that the series diverges.

Answer 2

Have you try \begin{align} \sum^\infty_{n=1}\frac{1}{\sqrt{n}} \leq \sum^\infty_{n=1}\frac{5^n}{\sqrt{n}4^n} \end{align}