<p>Does the Calculus BC exam require telescoping series knowledge and root test? I only see p series, geometric, nth term, ratio, alternating, and maybe direct comparison (usually for comparison with the harmonic series). </p>
<p>And I don’t really understand the nth term test, if the limit as An goes to infinity is not zero, can’t it still converge to another number like 1? Or am I just totally wrong and should just go with the if the limit to infinity is not zero then it diverges?</p>
<p>I don’t think you need to know the telescoping series test or the root test for the AP Exam as the other tests are enough to find convergence or divergence. And as for the nth term test, if it is 0, the series diverges. If it isn’t 0, the test is inconclusive and you have to try another test.</p>
<p>Telescoping is just a neat one to know, because you can find the sum really really quickly. And indeed, you can use ratio test en lieu of root test everywhere it pops up (although root test is usually somewhat faster). I don’t think I’ve seen either of those on the AP exam, and I’m admittedly a little too lazy to look up the requirements.</p>
<p>You should know limit comparison (I see that missing from your list). That one’s handier than direct comparison and will almost definitely appear.</p>
<p>As far as the nth-term goes, they’re just saying that if the terms don’t go to 0, the series diverges.</p>
<p>To see how this works, let’s pretend it doesn’t for a second, and let the terms go to some other number, like 1.</p>
<p>Then, the sum of the terms really late on in the series looks like this:</p>
<p>1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + …</p>
<p>Of course, they’re likely not exactly one, but we get the idea.</p>
<p>So the question then becomes, what single number does the sum approach over time?</p>
<p>Well, the more terms we add on, the more the sum grows, and if we add an infinite number of these ones, we’ll eventually get infinitely large, which isn’t actually a number.</p>
<p>In fact, the only way to let the sum approach a single number in the long run is if we start adding numbers together that are so small that those numbers are practically zero.</p>
<p>And that’s really all the nth-term test says.</p>
<p>How can I find the sum if the series isn’t geometric and I don’t know the telescoping series sum?</p>
<p>So let’s take a simple telescoping series: sigma(n = 3 to infinity, [1/(n-2) - 1/n]).</p>
<p>Let’s just start by writing out a few terms:</p>
<p>When n = 3, we have 1 - 1/3. (I have intentionally not simplified this.)
When n = 4, we have 1/2 - 1/4.
When n = 5, we have 1/3 - 1/5.
When n = 6, we have 1/4 - 1/6.
When n = 7, we have 1/5 - 1/7.
When n = 8, we have 1/6 - 1/8.
When n = 9, we have 1/7 - 1/9.
When n = 10, we have 1/8 - 1/10.</p>
<p>Notice that each of these terms has two parts to it, a positive part and a negative part. With two exceptions, each negative part has a corresponding positive part two terms later. Those exceptions are the positive portions of 1 and 1/2 (from n = 3 and n = 4) and the negative portions of 1/(some really huge number) and 1/(some really huge number + 1) that occur at the “end” of this infinite series.</p>
<p>Of course, as we continue to work our way further out, 1/(some really huge number) approaches 0, and so the negative portions tend to go towards zero. Accordingly, the sum is 1 + 1/2 = 3/2.</p>
<p>What if the terms are neither geometric nor telescoping? Then I would guess I’d make an improper integral, but I have no idea how to integrate a factorial…</p>
<p>Most of the time, they don’t ask you to find the sum in a case like that (or it’s simple enough that you can manually add the initial terms for a partial sum), just to verify that the series converges and that such a sum exists.</p>
<p>One caution is that integrating doesn’t necessarily do the trick. The Integral Test is actually quite specific to say that either the improper integral and the series both converge or both diverge, but does not actually say that when both converge, they actually converge to the same number. In fact, cases exist where they don’t.</p>
<p>How do I know if a Taylor series is an over/under approximation of another point? Like if the Taylor series is for x=2 but we’re approximating x=1.9?</p>
<p>Nevermind, if the Taylor Series is only up to the 1st term, it is a tangent haha</p>