9.3 Integral Test and P-Series

p-series Test converges if, diverges if. We could show this with the integral test.

the harmonic series: diverges. (It is a p-series with p=1.) It diverges very slowly, but it diverges. Because the p-series is so easy to evaluate, we use it to compare to other series.

Use a P-Series test to determine if the series appears to converge a)¼ + 1/9 + 1/16+ 1/25+ … _ _ _ b) 1 + 1/√4 + 1/√9 + 1/√25 + …

Use a P-Series test to determine if the series appears to converge ∞ a)¼ + 1/9 + 1/16+ 1/25+ … = Σ 1/n 2 ∞ n=2 Σ(1/n) 2 Is a p series with p= 2 converges n=2 _ _ _ ∞ b) 1 + 1/√4 + 1/√9 + 1/√25 + … = Σ 1/(n) 1/2 n=1 Is a p series with p = ½ diverges

Remember that when we first studied integrals, we used a summation of rectangles to approximate the area under a curve: This leads to: The Integral Test If is a positive sequence and where is a continuous, positive decreasing function, then: and both converge or both diverge.

Example 1: Does converge? Since the integral converges, the series must converge. (but not necessarily to 2.)

Flow chart (so far) Does a series converge? Does the nth term approach zero → no diverges If Yes or maybe Is it a special series Geometric series →yes converges if │r│ < 1 to a/(1- r) P series →yes converges if p >1 Telescoping series →yes converges to uncancelled terms If No Use a general test The only general test that we have so far is the Integral test The series converges if the corresponding integral converges (only works for series that are positive, continuous and decreasing)

Use any method learned so far to determine if a series converges or diverges ∞ ∞ ∞ Σ (1/n.95 )Σ (1/n 2 - 1/(n+1) 2 )Σ x/lnx n=1 n=1 n=1 ∞ Σ4n/(2n 2 +1)Σ (n+1)/(n+5) n=1

Use any method learned so far to determine if a series converges or diverges ∞ ∞ ∞ Σ (1/n.95 ) Σ (1/n 2 - 1/(n+1) 2 ) Σ x/lnx n=1 n=1 n=1 DivergesConvergesDiverges P-seriesTelescoping nth term test ∞ Σ4n/(2n 2 +1)Σ (n+1)/(n+5) n=1 Diverges Integral testnth term test

p odd,17, 27-41odd, odd