Is the alternating series $\sum_{n=1}^{\infty }(-1)^n \frac {n^2 - 1}{2n^2 + 3}$ divergent?

Question

I tried every test for convergence and really came up with nothing. Without answering the problem directly, is it possible to determine the divergence or convergence for this series?

$$\sum_{n=1}^{\infty }(-1)^n \frac {n^2 - 1}{2n^2 + 3}$$

If it can be determined could someone give a little hint as to what direction to take? Thank you!

Answer 1

A necessary condition for any convergent series is that its terms sequence converges to zero... This must be true for all convergent series : positive, alternating, whatever.

In your case, we have

$$\lim_{n\to\infty}\frac{(-1)^n(n^2-1)}{2n^2+3}=\lim_{n\to\infty}\frac{(-1)^n\left(1-\frac1{n^2}\right)}{2+\frac3{n^2}}\ldots\text{doesn't exist at all}$$

Answer 2

Let $$a_n=(-1)^n\frac{n^2-1}{2n^2+3}$$

we have $$\forall n\in \mathbb N^* \;\; |a_n|=\frac{1-\frac{1}{n^2}}{2+\frac{3}{n^2}}$$

which yields to

$$\lim_{n\to +\infty}|a_n|=\frac{1}{2}$$

and

$$\lim_{n\to+\infty} a_n \neq 0$$

thus, the series $\sum a_n$ is divergent.