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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Professor Leonard, thank you for a fantastic explanation of the Integral and P-Series Test for Convergence/Divergence in Calculus Two. The P-Series Test is simple to follow and understand, however showing a function is decreasing to satisfy the Integral Test is problematic. Many Calculus textbooks does not show this important step in their Integral Test examples.

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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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Can anyone help me at around 19:10? Professor Leonard says that he "doesn't know" whether the limit of the integral is equal to the sum of the series, but doesn't the fact that it is PCD pretty much guarantee that it is not the sum? Is there a counter example to my hypothesis?

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For the last example would you be able to separate it into the sum from 1 to infinity for e^(1/n) times the sum from 1 to infinity of 1/(n^2) and evaluate each of them separately and just multiply them in the end?

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I don't understand how at 13:45 , he asks if it's positive on 1 to infinity but if x were to be 1 then wouldn't that make the denominator 0 thus making the function undefined?

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On ln n /n integration, whats the reason to not set the range equal to [e,infinity) instead of [3,infinity) ? seems like it will cancel out ln and make the question easier, although it wasnt really needed since the first term was infinity.

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For the p test what if p is 1.00001 does it converge? Still greater than 1, but not by much

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why does the divergent test not work on example at 26:40? Professor says it is convergent on 0 if you take the limit. But by the end of the problem it is divergent?

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It's because the divergence test states that @ lim=0 it may converge, but it's not guaranteed.

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27:30 do we need to look for positiveness or non-negativeness?

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For the series from 1 to infinity of 1/(n^2+1), couldn't you also use the Direct Comparison Test and compare it with 1/n^2, which is a convergent P-series?

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By the way, the requirements for the function can be relaxed to non-increasing and non-negative instead of continuous, decreasing and positive.

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