Calculus 2 Lecture 9.3: Using the Integral Test for Convergence/Divergence of Series, P-Series

Calculus 2 Lecture 9.3: Using the Integral Test for Convergence/Divergence of Series, P-Series

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[shawnarmstrong]

00:00:00 INFO: Integral Test Explanation
*) f(x) = f(n) = a_n
*) f(n) must be continuous, positive & decreasing
*) If integral converges, so will series
*) Does not always provide sum of series
00:10:36 INFO: Integral Test Example 1
*) Step 1: Identify f(x)
*) Step 2: Determine if it is positive, continuous & decreasing
*) Step 3: Solve improper integral
*) Step 4: Connect the dots: lim f(x) exist ∵ series is convergent
00:19:40 INFO: Integral Test Example 2
00:26:27 INFO: Integral Test Example 3
*) Proving f(x) = ln(x)/x is decreasing
*) The integral diverges ∵ series is divergent
00:41:08 INFO: P Series Explanation
*) P series will always diverge when P <= 1
*) P series will converge when P > 1
00:47:20 INFO: P Series Summary
00:49:38 INFO: P Series Example 1
00:51:00 INFO: P Series Example 2
*) Root P
00:52:00 INFO: P Series Example 3
*) Negative P
00:57:45 INFO: Integral Test Example 4
*) Will divergence test work? No
*) Is it a geometric series? No
*) Is it a p series? No
*) Can you use Integral Test? Yes, f(x) is positive, continuous & decreasing

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[ProfessorLeonard]

Raquel Gallo Hello, I didn't elaborate because this particular example was covered in section 7.6 at 43:15.  Please check it out!  Basically, the divergence test for the example you are referencing here shows divergence for any "p" = 0 or "p" < 0.  For "p" = 1, we create a Harmonic Series (Divergent).  For "p" between 0 and 1, we use the example of an improper integral to show that the function of 1/x^p (I believe I use x^n in 7.6) is Divergent for "p" between 0 and 1.  Hence, by the Integral Test, the series from 1 to inf.  of 1/n^p must also be Divergent on this interval.  Sorry for the confusion, but we had covered this in class at a previous time in a section on integrals.  Cheers and thanks for watching!!
Professor Leonard

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