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

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!

• Thats because the series is similar to a geometric series, so the root, or better the ratio test should be used. – Rene Schipperus Apr 24 '18 at 23:39
• Could have been done by nth term test for divergence actually – chhro Apr 24 '18 at 23:42
• MathJax works in the title, don't you know? – Shaun Apr 24 '18 at 23:53

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 $a_n$.
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.