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Home , Absolute convergence, Conditional convergence

MA 146 Steve Sawin

11.6 and the

Absolute and Last time we distinguished between the terms ak of a , and the positive version of those terms, which for we called bk but in general are just |ak|, the absolute values of the terms. We say that a series is Absolutely Convergent if the series made up of the absolute values |ak| of the original terms is convergent. This ∞ X (−1)k k2 k=1 is absolutely convergent because ∞ X 1 k2 k=1 converges (PST with p = 2) and ∞ X (−1)k k k=1 is not absolutely convergent because ∞ X 1 k k=1 diverges (PST with p = 1). The first key fact about absolute convergence is that P Absolute Convergence Test If a series ak is absolutely convergent then it is con- vergent. This makes sense because changing the sign on some of the terms is only going to cancel out some of the growth of the partial sums and therefore make it easier to converge. If a series converges but is not absolutely convergent, we say it is conditionally con- vergent. So is exactly one of the three choices absolutely convergent/conditionally convergent/divergent. Examples: Are the following series absolutely convergent, conditionally convergent or divergent?

∞ X (−1)k−1 • k2 k=1 ∞ X (−1)n • n n=1 Answers: For the first one, try first if it is absolutely convergent. The series is ∞ X 1 . k2 k=1 This a p-series with p = 2, so by the PST it is convergent. So the original series is absolutely convergent, we need do no more. For the second series, the absolute version is

∞ X 1 n n=1 which is a p-series with p = 1 so by PST it is divergent. That means the original series can be conditionally convergnt or divergent. The only thing that will decide that for us is the AST. It is an alternating series and the positive version of the terms 1/n are clearly decreasing, so since lim 1/n = 0 n→∞ by the AST the original series is conditionally convergent.

∞ X (−1)k+1 • 3k k=0 ∞ X (−2)k • k3 + 1 k=1 ∞ X (−1)nn2 • . n3 + n + 2 n=1 The reason we care about absolute versus conditional convergence is technical. There are certain things you can do with finite sums that you would like also to do with infinite sums. For instance if you take the derivative of a sum of functions, the derivative of the sum is the sum of the derivative. Believe it or not, we will want that to be true of an infinite sum of functions! This, and lots of similar things, work as long as the sum is absolutely convergent. Here is one cool example. P∞ k+1 We know that the alternating harmonic series k=1(−1) /k is conditionally conver- gent. What does it converge to? We will see later in the semester that it converges to ln(2) 1 1 1 1 1 ln(2) = 1 − + − + − + ··· 2 3 4 5 6 But, if you rearrange the terms of this sum into a different order you can make it converge to whatever you like!. Rearranging terms in a finite sum does not change the result (com- mutativity of addition), but in an infinite sum it can, unless you are in the happy situation of absolute convergence. Using the Ratio Test The LCT (Limit Comparison Test) is a really precise tool, that always works as long as the simplifed series is p-series or a geometric series. If you have things in your term like k! or k2k or ln(k), this may not happen. For these situations the ratio test, may work, although it m,isses a lot of subtlety. Here is the idea. In a geometric series, each term is r times the previous, and if |r| < 1 the terms are getting small so fast that the sum converges, otherwise it diverges. If you look at the ratio of successive terms of your series and that ratio gets very close to some constant r as k gets large, then it will behave more or less like a geometric series with that r, and either converge or diverge according to the Geometric Series Test. The care statement is Ratio Test: If the absolute value of the ratio of successive terms of a series approaches a number r as the index goes to ∞

ak+1 lim = r k→∞ ak then  Converges Absolutely if r < 1  ∞  X Diverges if r > 1 ak  k=0  if r = 1  Warmup with silly examples: First we will do three examples that we can do by other means. This will give us a feel for how the calculation goes. Let’s see what the ratio test says about the convergence of the following three series

∞ X 1 • 3k k=0 Answer: We always put the term of the original series in the denominator, and we put the same formula but with every k replaced by k + 1 in the numerator. This is

k+1 1/3 lim . k→∞ 1/3k Like in LCT, we always will start this process by flipping the denominator over. Also, everything is positive, so we can erase the absolute value signs. This gives us

3k = lim . k→∞ 3k+1 Now we need ... wait for it... the rules of exponents! Same base, different exponents, so subtract exponents to get 1 1 = lim = . k→∞ 3 3 Savor this, because the same thing will always happen with exponential terms: subtracting exponents makes the k disappear and we are just left with the base if the exponent. We can now apply the RT to conclude 1 r = < 1 =⇒ Absolutely Convergent. 3 You can think of this calculation as telling you the r of the Geometric Series which your series behaves most like. Of course, our series is already Geometric with r = 1/3 so of course we got 1/3 out.

∞ X 1 • k2 k=1 Answer: We always put the term of the original series in the denominator, and we put the same formula but with every k replaced by k + 1 in the numerator. This is

2 1/(k + 1) lim . k→∞ 1/k2 Notice what replacing k with k + 1 looks like here. Flip the denominator over, erase the absolute value signs. This gives us k2 = lim . k→∞ (k + 1)2 Here we can just do our usual limit finding tricks, throwing away the lower order terms (which is just the +1 in the denominator) to get k2 = lim = 1. k→∞ k2 Savor this, because the same thing will always happen with polynomial terms: A big messy ratio will just turn out to be 1.

r = 1 =⇒ .

The original series did not really look like any geometric series, so we got no value out of the test.

∞ X 1 • k! k=1 Answer: Original terms in the denominator, same but with every k replaced by k + 1 in the numerator 1/(k + 1)! lim . k→∞ 1/k! Flip denominator, erase the absolute value signs k! = lim . k→∞ (k + 1)! Now Remember the recursive definition of the ! we know that (k + 1)! = (k + 1)k!. That means that (k + 1)!/k! = k + 1 and k!/(k + 1)! = 1/(k + 1). So 1 = lim = 0. k→∞ k Savor this, because the same thing will always happen with factorial terms: The recursive trick for will make it into a very simple limit which will generally go to 0 or infinity. r = 0 =⇒ Absolutely Convergent. Because k! grows faster than any exponential, 1/k! gets smaller than any exponential, so it sum is smaller than any geometric series, so it acts as if r = 0. Everything else puts these three cases together somehow. Examples: Determine if the following series are absolutely convergent or divergent with the ratio test ∞ X k2 • 2k k=0 ∞ X k + 1 • k3 + 1 k=0 ∞ X 2k • k! n=0 Answer 2 k+1 k 2 (k + 1) /2 2 (k + 1) lim = lim k→∞ k2/2k k→∞ k22k+1 2k(k + 1)2 2k (k + 1)2 = lim = lim lim k→∞ 2k+1k2 k→∞ 2k+1 k→∞ k2 1 k2 1 = lim = < 1 2 k→∞ k2 + 2k + 1 2 so the series is absolutely convergent by the RT. Notice we collected exponentials and polynomials and took their limits separately, modeling on the three previous examples. Answer 3 3 ((k + 1) + 1)/((k + 1) + 1) (k + 1)(k + 2) lim = lim k→∞ (k + 1)/(k3 + 1) k→∞ (k + 1)((k + 1)3 + 1) k · k3 = lim = 1 k→∞ k · k3 so we learn nothing. If all you see in the original problem is polynomials and square roots, this is how it will always go, so only use RT when you see big ticket items like 2k and k! in your terms. Answer

k+1 k+1 2 /(k + 1)! k!2 lim = lim k→∞ 2k/k! k→∞ 2k(k + 1)! 2k+1 k! = lim = lim k→∞ 2k k→∞ (k + 1)! 1 = 2 lim = 0 < 1 k→∞ k + 1 and the series converges. More examples: Use RT to decide which series converge and diverge

∞ X 3k • k22k k=1 ∞ X k25k • k! k=1