# Solution Verification using the Divergence Test

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

• 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. Apr 20, 2021 at 19:37
• 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 $1/e$. Apr 20, 2021 at 19:38

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.