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

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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 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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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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Professor in 38:50 I used integration by parts. I got ln(x^lnx)*1/2. Is that right?

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Just a quick note at 58:03 -- the limit is indeed 0 but you can't use l'Hopital's. The numerator e^(1/n) goes to 1 and the denominator n^2 goes to infinity so you simply get 0.

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at minute 41:00 aren't we supposed to substitute 3 and b in our U that we took to get our new bound or interval?

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The divergence test doesn't show that 1/n^p is divergent when 0

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