How does this series diverge by limit comparison test? $$\sum_{n=1}^\infty \sqrt{\frac{n+4}{n^4+4}}$$

I origionally tried using $\frac{1}{n^2}$ for the comparison, but I'm pretty sure it has to be $\frac{n}{n^2}$ to properly compare.


2 Answers 2


$\frac{n+4}{n^4+4} \leq \frac{n+4n}{n^4} \leq \frac{5}{n^3}$,

hence, $\sqrt{\frac{n+4}{n^4+4}} \leq \frac{\sqrt{5}}{n^{3/2}}$

so the series converges by comparison with convergent p-series $\sum \frac{1}{n^{3/2}}$

  • $\begingroup$ Wont c be equal to 0 though once you take the limit? c must be greater than 0 to conclude the test with b sub n $\endgroup$
    – Luke D
    Dec 18, 2018 at 5:41
  • $\begingroup$ @LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = \sqrt{5}$. $\endgroup$ Dec 18, 2018 at 5:43
  • $\begingroup$ This can also be done with the limit comparison test, which is what the OP asked for. $\endgroup$
    – David
    Dec 18, 2018 at 5:45
  • $\begingroup$ Yeah im curious to see the limit comparison test. But regardless, thanks for the soln. $\endgroup$
    – Luke D
    Dec 18, 2018 at 5:48

For this sort of thing it is strongly advised to do a rough calculation first. We have $$\sqrt{\frac{n+4}{n^4+4}}\approx\sqrt{\frac{n}{n^4}}=\frac1{n^{3/2}}\ ,$$ which suggests comparing with $$\sum\frac1{n^{3/2}}\ .$$ We have $$\sqrt{\frac{n+4}{n^4+4}}\bigg/\frac1{n^{3/2}}=\sqrt{\frac{n^4+4n^3}{n^4+4}} =\sqrt{\frac{1+4n^{-1}}{1+4n^{-4}}}\to1\quad\hbox{as $n\to\infty$}\ .$$ Since this limit exists and is finite and not zero, and we know that $$\sum\frac1{n^{3/2}}$$ converges, your series converges too. (Doesn't diverge!!!)

  • $\begingroup$ So the answer our teacher gave us is wrong... thanks for the heads up. $\endgroup$
    – Luke D
    Dec 18, 2018 at 5:46
  • $\begingroup$ If your teacher said the series diverges, yes, that's wrong. $\endgroup$
    – David
    Dec 18, 2018 at 5:47

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