Limit Comparison Test for Convergence of an Infinite Series
Key Questions
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Well, I would try to see if I can directly compare first; however, it might not be easy when its expression is complicated. The benefit of the limit comparison test is that we can compare series without verifying the inequality we need in order to apply the direct comparison test, of course, at the cost of having to evaluate the limit.
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Let
#a_n={n^2-5n}/{n^3+n+1}# .By using the leading terms of the numerator and the denominator, we can construct
#b_n={n^2}/{n^3}=1/n# .Remember that
#sum_{n=1}^infty b_n# diverges since it is a harmonic series.By Limit Comparison Test,
#lim_{n to infty}{a_n}/{b_n}=lim_{n to infty}{n^2-5n}/{n^3+n+1}cdot n/1 =lim_{n to infty}{n^3-5n^2}/{n^3+n+1}# by dividing the numerator and the denominator by
#n^3# ,#=lim_{n to infty}{1-5/n}/{1+1/n^2+1/n^3}={1-0}/{1+0+0}=1 < infty# ,which indicates that
#sum_{n=1}^infty a_n# and#sum_{n=1}^infty b_n# are comparable.Hence,
#sum_{n=1}^infty{n^2-5n}/{n^3+n+1}# also diverges.
I hope that this was helpful.
Questions
Tests of Convergence / Divergence
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Geometric Series
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Nth Term Test for Divergence of an Infinite Series
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Direct Comparison Test for Convergence of an Infinite Series
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Ratio Test for Convergence of an Infinite Series
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Integral Test for Convergence of an Infinite Series
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Limit Comparison Test for Convergence of an Infinite Series
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Alternating Series Test (Leibniz's Theorem) for Convergence of an Infinite Series
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Infinite Sequences
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Root Test for for Convergence of an Infinite Series
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Infinite Series
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Strategies to Test an Infinite Series for Convergence
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Harmonic Series
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Indeterminate Forms and de L'hospital's Rule
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Partial Sums of Infinite Series