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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!

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

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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}

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  • $\begingroup$ how about trying LIMIT comparison instead of direct comparison? $\endgroup$ – NL628 Apr 24 '18 at 23:38

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