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

1 Answer
Ernest Z.
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 = 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.

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