I just wanted to check some of my solutions here concerning determining the convergence or divergence of a series using the divergence test. Can the divergence test be used to see if the following series diverges:

  1. $\sum_{n=1}^{\infty}(\frac{\pi}{3})^n$. The answer here is yes since the series diverges.
  2. $\sum_{n=1}^{\infty}(-1)^n \arctan(2n)$. The answer here is again, yes, since $\arctan(2n)$ converges to $\frac{\pi}{2}$, which means it diverges.
  3. $\sum_{n=1}^{\infty}\frac{n^2}{n^3+2}$. Sadly this one seems to not be applicable. It converges to 0, which means no information can be given.
  4. $\sum_{n=1}^{\infty}(\frac{n}{n+1})^n$. This would just be $(1)^n$ when $n$ approaches infinity. So it diverges.

Thank you for your time :D

  • $\begingroup$ Looks about right. You should be more careful distinguishing between the convergence/divergence of the series ($\sum_n a_n$) and of the sequence ($(a_n)_n$); you mean the right thing, but you need to state it properly. Also, in the second statement, you have to consider the $(-1)^n$ also in the argument; that does not change the outcome though. $\endgroup$
    – S.Farr
    Apr 20, 2021 at 19:37
  • 2
    $\begingroup$ Your answer for $(4)$ is right, but it's not necessarily true that $\lim_{n \to \infty} a_n^{b_n}\not=0$ if $a_n \to 1$ and $b_n \to \infty$. In this case the summands converge to $1/e$. $\endgroup$ Apr 20, 2021 at 19:38

1 Answer 1


You're right. Except for the third one, the divergence test applies, since the general terms don't go to zero. (As noted in the comments, $1/e $ is the actual limit for the last one.)

Btw, $3$ also diverges, by the comparison test. Namely $\dfrac {n^2}{n^3+2}=\dfrac1 {n+2/n^2}\ge\dfrac1 {n+2} $. And $\sum\dfrac1 {n+2} $ diverges, as it is just the harmonic series minus the first two terms.


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