Home > Calculus > Direct Comparison Test for Convergence of an Infinite Series

How do you use the direct comparison test for infinite series?

Answer 1

If you are trying determine the conergence of #sum{a_n}#, then you can compare with #sum b_n# whose convergence is known.

If #0 leq a_n leq b_n# and #sum b_n# converges, then #sum a_n# also converges. If #a_n geq b_n geq 0# and #sum b_n# diverges, then #sum a_n# also diverges.

This test is very intuitive since all it is saying is that if the larger series comverges, then the smaller series also converges, and if the smaller series diverges, then the larger series diverges.

Let us look at some examples.

Example 1: #sum_{n=1}^{infty}{sin n+1}/{n}# Since #{sin n+1}/n geq 1/n# and #\sum_{n=1}^{infty}1/n# is the harmonic series, which is divergent, we may conclude that #sum_{n=1}^{infty}{sin n+1}/{n}# is also divergent by the direct comparison test.

Example 2: #sum_{n=1}^{infty}{cos^2n}/n^{3/2}# Since #{cos^2n}/{n^{3/2}} leq 1/n^{3/2}# and #sum_{n=1}^{infty}1/n^{3/2}# is a convergent p-series (p>1), we may conclude that #sum_{n=1}^{infty}{cos^2n}/n^{3/2}# is also convergent by the dirct comparison test.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.