Question

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

Answer 1

`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 = 1/(4n^2-1)`

Let's think about the end behaviour of `a_n`.

For large `n`, the denominator `4n^2-1` acts like `4n^2`.

So, for large `n`, `a_n` acts like `1/(4n^2)`.

Let `b_n= 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= 1/n^2` is convergent, so

`a_n = 1/(4n^2-1)` is also convergent.