Comparision test for this series?

Question

How do I check divergence of this series?

$$\sum_{n=0}^{\infty} \frac{6}{4n-1} - \frac{6}{4n+3}$$

Wolframalpha said it used the comparision test but I don't see what possible smaller sum to use?

It's also a telescoping series but I don't know how that is useful in this case?!

Answer 1

$$\sum_{n=0}^{\infty} \frac{6}{4n-1} - \frac{6}{4n+3}=24\sum_{n=0}^{\infty}\frac{1}{(4n-1)(4n+3)}$$ Now as $4n-1>n$ f0r $n>0$ and $4n+3>n$ so $$\frac{1}{(4n-1)(4n+3)}\leq \frac{1}{n^2}$$

Answer 2

$\sum_{n=0}^m(6/(4 n-1)-6/(4 n+3)) = -6/(4 m+3)-6$ where m is finite for the rearranging, but now we let it go $\to \infty$. What happens?