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

Question

Ezekiel Alexander · Accepted Answer

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

#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.

Isabella Ackerman · Answer

undefined To use the Comparison Test to determine the convergence of the series $ \sum \frac{1}{4n^2 - 1} $ as $ n $ approaches infinity:

1. Consider the series $ \sum \frac{1}{4n^2} $.
2. Observe that $ \frac{1}{4n^2} $ is a term of the harmonic series, which is known to diverge.
3. Note that $ 4n^2 - 1 > 4n^2 $ for all positive integers $ n $.
4. Utilize the Comparison Test, stating that if $ a_n $ is less than or equal to $ b_n $ for all $ n $ and $ \sum b_n $ converges, then $ \sum a_n $ converges.
5. Since $ \frac{1}{4n^2 - 1} < \frac{1}{4n^2} $ for all positive integers $ n $, and $ \sum \frac{1}{4n^2} $ diverges, $ \sum \frac{1}{4n^2 - 1} $ must also diverge.