Question regarding a series that contains logarithms $$\sum_{n=1}^\infty\bigl(3\log(n^2+1)-2\log(n^3+1)\bigr)$$
I tried the limit comparison test with $\sum_{n=1}^\infty\frac1{n^2}$ and then applied L'Hopital's Rule. What I ended up was a limit which was a real number and as the series of $\frac1{n^2}$ converges, my original series converges as well. However, I was wondering if there is another way of solving the problem, one that involves fewer computations.

$$3\ln(n^2+1)-2\ln(n^3+1)=\ln\left( \frac{(n^2+1)^3}{(n^3+1)^2}\right)=\ln\left( \frac{n^6+3n^4+3n^2+1}{n^6+2n^3+1}\right)$$
$$=\ln\left( 1+\frac{3n^4-2n^3+3n^2}{n^6+2n^3+1}\right) \sim \frac{3n^4-2n^3+3n^2}{n^6+2n^3+1} \sim \frac{3}{n^2}$$
Therefore the series converges.
