Convergence of $\sum_{n=0}^{+ \infty} \frac{3}{2n² -1}$ 
Does $$\sum_{n=0}^{+ \infty} \frac{3}{2n² -1}$$
converge?

I did the following:
$$\frac{3}{2n²-1} = \frac{3}{2n²}\left(\frac{1}{1-\frac{1}{2n²}}\right)$$
and because $$\frac{1}{1-\frac{1}{2n²}} \to 1$$
the sequence is bounded, and hence, there exists $M \in \mathbb{R}^{+}$
such that:
$$\left|\frac{1}{1-\frac{1}{2n²}}\right| = \frac{1}{1-\frac{1}{2n²}} < M$$
Hence:
$$\frac{3}{2n²-1} < \frac{3M}{2n²}$$
and because $\sum n^{-2}$ converges, by the comparison test it follows that the given sequence converges.
Is this correct? Is there an easier approach?
Thanks in advance.

How about
$$
    \frac{3}{2n^2-1} \leq \frac{3}{n^2}
$$
for all $n \geq 1$.
