# How do you test the series Sigma (3n^2+1)/(2n^4-1) from n is [1,oo) for convergence?

Feb 10, 2017

${\sum}_{n = 1}^{\infty} \frac{3 {n}^{2} + 1}{2 {n}^{4} - 1}$

is convergent based on the direct comparison test.

#### Explanation:

We can test the series by direct comparison. As:

${\sum}_{n = 1}^{\infty} {a}_{n} = {\sum}_{n = 1}^{\infty} \frac{3 {n}^{2} + 1}{2 {n}^{4} - 1}$

is a series with positive terms, we need to find a convergent series ${\sum}_{n = 1}^{\infty} {b}_{n}$ such that:

${a}_{n} < {b}_{n}$ for $n > N$

Now we have:

$\frac{3 {n}^{2} + 1}{2 {n}^{4} - 1} < \frac{3 {n}^{2}}{2 {n}^{4} - 1} < \frac{3 {n}^{2}}{2 {n}^{4}} < \frac{1}{n} ^ 2$

But ${\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 2$ is convergent based on the p-series test, so also:

${\sum}_{n = 1}^{\infty} \frac{3 {n}^{2} + 1}{2 {n}^{4} - 1}$

is convergent