# How do you use the Comparison Test to see if 1/(4n^2-1) converges, n is going to infinity?

Aug 11, 2015

color(red)(sum_(n=1)^∞ 1/(4n^2-1)" is convergent").

#### Explanation:

sum_(n=1)^∞ 1/(4n^2-1)

The limit comparison test states that if ${a}_{n}$ and ${b}_{n}$ are series with positive terms and if lim_(n→∞) (a_n)/(b_n) is positive and finite, then either both series converge or both diverge.

Let ${a}_{n} = \frac{1}{4 {n}^{2} - 1}$

Let's think about the end behaviour of ${a}_{n}$.

For large $n$, the denominator $4 {n}^{2} - 1$ acts like $4 {n}^{2}$.

So, for large $n$, ${a}_{n}$ acts like $\frac{1}{4 {n}^{2}}$.

Let ${b}_{n} = \frac{1}{n} ^ 2$.

Then lim_(n→∞)a_n/b_n = lim_(n→∞)(1/(4n^2-1))/(1/n^2)= lim_(n→∞)n^2/(4n^2-1) = lim_(n→∞)1/(4-1/n^2) = 1/4

The limit is both positive and finite, so either ${a}_{n}$ and ${b}_{n}$ are both divergent or both are convergent.

But ${b}_{n} = \frac{1}{n} ^ 2$ is convergent, so

${a}_{n} = \frac{1}{4 {n}^{2} - 1}$ is also convergent.