# 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?

How about $$\frac{3}{2n^2-1} \leq \frac{3}{n^2}$$ for all $n \geq 1$.
Your series is absolutely convergent by comparison with $\sum_{n\geq 1}\frac{3}{n^2}=\frac{\pi^2}{2}$, for instance.
You may also compute it in a explicit way. If we start with $$\frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right)$$ and apply $\frac{d}{dz}\log(\cdot)$ to both sides, we get $$-\frac{1}{z}+\pi\cot(\pi z) =\sum_{n\geq 1}\frac{2z}{z^2-n^2}$$ and by evaluating both sides at $z=\frac{1}{\sqrt{2}}$ $$1-\frac{\pi}{\sqrt{2}}\,\cot\left(\frac{\pi}{\sqrt{2}}\right)=\sum_{n\geq 1}\frac{1}{n^2-\frac{1}{2}}$$ so: $$\boxed{ \sum_{n\geq 1}\frac{3}{2n^2-1} = \color{red}{\frac{3}{2}-\frac{3\pi}{2\sqrt{2}}\,\cot\left(\frac{\pi}{\sqrt{2}}\right)}\approx 4.036518.}$$
Remember higher degree term of polynomial dominates the whole polynomial So,2$n^{2}$-1~$2n^{2}$..can you use limit comparison test now..