How can I find out if this sum converges or diverges?

I believe that is a question quite simple.

Until I looked in the documentation, but I think that I did not know how to do this.

Sum[(8 n + Sqrt[n])/(n^4 - n^2 + 5), {n, 1, Infinity}]

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I believe that is not only insert the sum. But i have to use another feature that i do not know what it was.

  • 5
    $\begingroup$ NSum[(8 n + Sqrt[n])/(n^4 - n^2 + 5), {n, 1, Infinity}] -> 3.50589 $\endgroup$
    – user36273
    Nov 26, 2016 at 11:18
  • $\begingroup$ Related: (126839). $\endgroup$
    – corey979
    Nov 26, 2016 at 16:31

2 Answers 2


This should do it:

SumConvergence[(8 n + Sqrt[n])/(n^4 - n^2 + 5), n]

It returns True.


When n is very large, it is like 1/n^3, it is convergence.

Series[(8 n + Sqrt[n])/(n^4 - n^2 + 5), {n, \[Infinity], 2}]

with reutern $O\left(\left(\frac{1}{n}\right)^3\right)$, so it is convergence.

  • $\begingroup$ Your answer is not related to Mathematica. $\endgroup$ Nov 26, 2016 at 19:41

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