11.3a: Positive-Term Series Integral test. P-series. Rita Korsunsky

Positive-term series are the series an such that an > 0 for every n. If an is a positive-term series and if there exists a number M such that: Sn = a1 + a2 + … + an < M for every n, then the series converges and has a sum S  M. If no such M exists, the series diverges.

The Integral Test: Sometimes we need to prove the right to left statements In using the integral test it is necessary to consider If f(x) is not easy to integrate, a different test for convergence or divergence should be used. When we use the Integral Test it is not necessary to start the series or the integral at . For instance, in testing the series Also, it is not necessary that be always decreasing. Instead, could be decreasing for larger than some number . Then is convergent, so is convergent too.

Total area of other rectangles < Exclude the first rectangle. Total area of other rectangles < area under the curve or Integral converges Partial sums are bounded =1  Series converges partial sums  1.64 as n   n Total area of all rectangles > area under the curve or Integral diverges Partial sums are unbounded  Series diverges partial sums   as n  

Let’s try! Why don’t we use inscribed rectangles Total area of all rectangles > area under the curve or Integral diverges Partial sums are unbounded  Series diverges partial sums   as n   Why don’t we use inscribed rectangles instead of circumscribed in 2nd problem? Let’s try! 1 2 3 4 Does it converge or diverge? Integral diverges So we can not make any conclusion. Use the circumscribed rectangles instead!

The series diverges by the integral test. Example 1 Prove that the harmonic series diverges using the integral test. Harmonic series: f > 0, continuous, and decreasing for x  1, so use the integral test: The series diverges by the integral test.

Converges But sum Example 2 f(x) > 0 and continuous if x  1. Does the infinite series converge or diverge? SOLUTION: f(x) > 0 and continuous if x  1. f(x) is decreasing Converges But sum

P-series (hyperharmonic series) where p > 0 Proof Use Integral test.

ILLUSTRATION: p-series value of p conclusion p = 2 Converges, p=2 > 1 p = 1/2 Diverges, p=½ < 1 p = 3/2 Converges, p=3/2 > 1 p = 1/3 Diverges, p=1/3 < 1