Divergent test for a simple series

Question

how would I be able to prove that, using the comparison test,

diverges?

Using symbolab gave me diverges, but it does not show how, and it used the series root test, which I will not cover in my course.

Thank you.

Answer 1

(Limit) Comparison test with $\sum_{n=1}^\infty \frac{1}{n}$, which diverges.

Answer 2

$\dfrac{2}{3n+1} > \dfrac{2}{4n} = \dfrac{1}{2}\cdot \dfrac{1}{n}$. You can compare your series with the one on the right: $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n}$. Thus your actual question is how to show this harmonic series diverges. Let $a_n = \displaystyle \sum_{k=1}^n \dfrac{1}{k}$. You can show $a_n \to +\infty$. Compare this sequence with $b_n = \text{lnn}$. You can prove by integration or induction that $a_n > b_n$, and $b_n \to +\infty$, so $a_n \to +\infty$, and you are done.

Alternately: you can prove that: $a_{2^n} \geq 1 + \frac{n}{2} \to +\infty$. So:

$a_{2^n} \to +\infty$, this implies: $a_n \to +\infty$.