AP Calculus BC Topic 10.9: Determining Absolute or Conditional Convergence

Topic 10.9 key terms

alternating series: A series whose terms alternate in sign, written $\sum (-1)^{n+1} b_n$ or $\sum (-1)^n b_n$ with every $b_n > 0$ .
alternating series test: If $b_n > 0$ , $b_{n+1} \le b_n$ (from some $n$ on) and $\lim_{n\to\infty} b_n = 0$ , then $\sum (-1)^{n+1} b_n$ converges. It never proves divergence.
absolutely convergent: $\sum a_n$ converges absolutely if $\sum |a_n|$ converges. An absolutely convergent series converges, and every rearrangement of it has the same sum.
conditionally convergent: $\sum a_n$ converges conditionally if it converges but $\sum |a_n|$ diverges, like $1 - \tfrac12 + \tfrac13 - \cdots$ .
ratio test: Let $L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|$ . If $L < 1$ the series converges absolutely; if $L > 1$ (or $L = \infty$ ) it diverges; if $L = 1$ the test is inconclusive.
rearrangement: A series with the same terms in a different order. Rearranging an absolutely convergent series never changes its sum; a conditionally convergent series can be rearranged to have any sum (Riemann).
partial sum: $S_n = a_1 + a_2 + \cdots + a_n$ , the sum of the first $n$ terms: where the shuttle stands after $n$ rods. The series converges when $S_n$ has a finite limit.

Transcript

596 words · 1 parts

How far did it travel?

Back to our first shuttle, resting at about zero point six nine, less than one unit from home. How far did it actually travel to get there? Let’s take every rod it used and lay them end to end, all pointing the same way. The length of this chain is the total distance the shuttle moved.

That’s one plus a half plus a third plus a quarter, the harmonic series, and you know it diverges. That means the chain never stops growing, while the shuttle barely moves at all. The shuttle ends up less than a unit from home after an infinite trip, because the backward rods cancel so much of the forward ones. Now cut the rods by a different rule, one over n squared, so one, a quarter, a ninth and so on. This time the unfolded chain stops too, at about one point six four.

This dial cuts every rod to one over n to the p. Turn it until the unfolded chain stops, and watch the shuttle while you search.

These two behaviors have names worth knowing by heart. A series converges absolutely if the series of its absolute values converges, so the total distance is finite. It converges conditionally if the series converges but the series of absolute values diverges, like our first shuttle. If the series itself doesn’t converge, it simply diverges.

Back to the one-over-n-squared rods. Their chain stops at about one point six four, and the shuttle settles. That’s a theorem. If a series converges absolutely, then it converges. Every wiggle spends some of the travel budget. Once the shuttle has spent almost all of its one point six four, the moves it has left add up to almost nothing, so it has to settle.

Here’s where you need it. Take sine of n over n squared. Its signs come in runs, plus, plus, plus, then minus, minus, minus. They don’t take turns, so the alternating series test can’t touch it.

The size of sine of n is never more than one, so the size of each term is at most one over n squared, a convergent p-series. By direct comparison, our series converges absolutely, so it converges. One more trap. Alternating doesn’t mean conditional. One minus a quarter plus a ninth alternates, and its chain stopped at one point six four. That’s absolute.

Now you pick the tool. The signs of cosine of n over n cubed come in runs too. Which tool settles this series?

Let’s do the whole thing the way a free-response question wants it, for one minus one over root two plus one over root three, and so on. For this classification, let’s check the absolute values first. Dropping the signs gives the sum of one over n to the one half, a p-series with p equals one half, so it diverges.

That rules out absolute convergence, but the signs might still save it. The signs alternate, one over root n gets smaller as n grows, and it shrinks to zero, so the series converges by the alternating series test. Putting the two halves together, you’d write the series converges by the alternating series test, but the sum of one over root n is a divergent p-series, so it converges conditionally.

Now you classify one, one minus a third plus a fifth minus a seventh, and so on. This one’s famous. Historians credit Madhava of Sangamagrama, in southern India, with discovering around fourteen hundred that it adds up to pi over four. Leibniz found it again in Europe almost three hundred years later.

Topic 10.9: Determining Absolute or Conditional Convergence

Topic 10.9 key terms

Transcript

How far did it travel?

Unit 10: Infinite Sequences and Series