Is the series $$\sum_\limits{n=1}^\infty\frac{n}{n^3+1}$$ convergent or divergent?

My answer is the following. Could anyone tell me if it is correct?

Since $$0<\frac{n}{n^3+1}<\frac{1}{n^2}\;\;,\;\;\;\;\forall n\in\mathbb{N}$$ and the series $$\sum_\limits{n=1}^\infty\frac{1}{n^2}$$ is convergent,

by applying comparison test, we get that the series $$\sum_\limits{n=1}^\infty\frac{n}{n^3+1}$$ is convergent too.

  • 9
    $\begingroup$ That is correct. $\endgroup$
    – markvs
    Aug 21, 2020 at 5:46
  • $\begingroup$ That's fine, for a case such $\sum_\limits{n=2}^\infty\frac{n}{n^3-1}$ we can use LCT. $\endgroup$
    – user
    Aug 21, 2020 at 5:50
  • $\begingroup$ $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$ and that is equivalent to $\sum_{n=1}^\infty \frac{n}{n^3}$. But $\frac{n}{n^3}>\frac{n}{n^3+1}$ for any $n\ge1$. Hence the given series converges. $\endgroup$ Aug 21, 2020 at 6:04

2 Answers 2


Yes, your answer is correct.

All the terms in your series are positive and all the terms in your series are less than $\frac{1}{n^2}$ (for $n\in\mathbb N$).
Since $\sum_{n=1}^\infty\frac{1}{n^2}$ is convergent and all the terms in your series (and hence all the partial sums of your series) are always less than that series, you can conclude, by the comparison test, that $\sum_{n=1}^\infty\frac{n}{n^3+1}$ is also convergent.


Let me add also some simple way in such situations: if we have series with positive general members $a_n, b_n$ and $a_n \sim b_n$ when $n \to \infty$, then series converges and diverges simultaneously. In your case $$\frac{n}{n^3+1} \sim \frac{1}{n^2}$$

  • $\begingroup$ By the way, this is known as the limit comparison test. $\endgroup$ Dec 13, 2020 at 22:58
  • $\begingroup$ Michael Spivak - Calculus, pages 475-476 (2008) $\endgroup$
    – zkutch
    Dec 13, 2020 at 23:35

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