Showing That the Sum of the Infinite Series cos(n)/n ConvergesDate: 04/01/2008 at 02:10:57 From: Mike Subject: Test of convergence of a given series The question is if the sum of the series from n = 1 to infinity of (cos n)/(n) converges or diverges. I can't seem to find a test to see whether the series converges or diverges. Each test that I've tried doesn't seem to solve the problem. I've tried the direct comparison test, the limit comparison test, the ratio test, and the root test. The problem seems very complicated, and I am utterly stumped! Date: 04/12/2008 at 20:03:57 From: Doctor Vogler Subject: Re: Test of convergence of a given series Hi Mike, Thanks for writing to Dr. Math. That's a very good question, and a very hard question too. It turns out that none of the tests that you usually learn in high school will be quite sufficient, although there is one known as the Dirichlet test (not usually taught in high school) which is a generalization of the Alternating Series test (meaning that the A.S. test is a special case of the Dirichlet test) which *is* sufficient. It is described at Wikipedia: Dirichlet Test http://en.wikipedia.org/wiki/Dirichlet_test and proven using "partial summation" Wikipedia: Partial Summation http://en.wikipedia.org/wiki/Partial_summation which is an analog for series to the method of partial integration (or "integration by parts") from calculus. You can also prove that the series converges using partial summation directly, or you can convert the series to an integral and then prove that the integral converges. Showing that the integral converges is not terribly easy either, although this method allows you not only to prove that the series converges but also to find its sum. All of these methods, however, share one other feature, which is that you have to start by knowing how to sum a series like cos 1 + cos 2 + cos 3 + ... + cos k. So let's start there. The way to sum this series, or any series of the form k sum cos(nx) n=1 for some number x is to use Euler's equation. See Imaginary Exponents and Euler's Equation http://mathforum.org/dr.math/faq/faq.euler.equation.html You change cos(nx) into (1/2)(e^(inx) + e^(-inx)), and then you split the summation into a sum of two geometric series. The formula for geometric series works in complex numbers just as it does for real numbers (since it's really just a polynomial factorization). Then you convert the result back into a formula in trig functions instead of complex numbers. In case that makes you uneasy, if you happen already to know the answer, then you can simply prove using trig formulas and induction that k sin((k+1/2)x) - sin(x/2) sum cos(nx) = ------------------------ n=1 2 sin(x/2) This is important for applying the Dirichlet test because you need to apply the Dirichlet test with a_n = 1/n and b_n = cos(n), and you need to prove that the sum k sin(k + 1/2) - sin(1/2) sum cos(n) = ------------------------ n=1 2 sin(1/2) is bounded independently of k. Fortunately, since sin(k + 1/2) is always between -1 and +1, that means that the sum of cos(n) from n=1 to n=k is always between -1/(2 sin(1/2)) - 1/2 and 1/(2 sin(1/2)) - 1/2, no matter what k is. And that's enough to use the Dirichlet test to prove that the series converges. The other method is a little trickier, but here's one way to find the sum of the series you want. The idea is to use the fact that the integral of sin(nx) dx is cos(nx) int sin(nx) dx = ------- + C. n Of course, we have to figure out what the constant is, so we should choose convenient bounds on the integral, such as pi cos(n) (-1)^n int sin(nx) dx = ------ - ------ 1 n n I'll leave it to you to prove that cos(n*pi) = (-1)^n when n is an integer. So this means that we need to find the summation k cos(n) k (-1)^n pi sum ------ = sum ------ + int sin(nx) dx n=1 n n=1 n 1 k (-1)^n pi k = sum ------ + int ( sum sin(nx) ) dx n=1 n 1 n=1 This means that you don't actually need the sum of cos(n), but you need the sum k cos(x/2) - cos((k+1/2)x) sum sin(nx) = ------------------------ n=1 2 sin(x/2) but you can find and/or prove this just like for the sum of cos(n). That means that k cos(n) k (-1)^n pi cos(x/2) - cos((k+1/2)x) sum ------ = sum ------ + int ------------------------ dx. n=1 n n=1 n 1 2 sin(x/2) This is interesting because it converts a sum with variable number of terms (k terms) into a known sum (the sum on the right side is well known to converge to -ln(2) when k goes to infinity, since it equals the Taylor series for -ln(1+x) at x=1) plus a single integral on a finite interval (1 to pi), and the k only appears in the summand, not in the interval. Well, it turns out that the first part of this integral can be evaluated exactly pi cos(x/2) pi int ---------- dx = [ ln |sin(x/2)| ] 1 2 sin(x/2) 1 = ln sin(pi/2) - ln sin(1/2) = ln 1 - ln sin(1/2) = -ln sin(1/2). The second part of the integral can be shown to approach zero as k grows to infinity, which means that -ln(2) - ln(sin(1/2)) = 0.042019505825368961725798384... is the sum of cos(n)/n. So the only thing left to do is prove my claim that the second part of the integral is smaller in absolute value than a constant times 1/k. (It turns out that the difference between -ln(2) and the sum of (-1)^n/n from n=1 to n=k is also smaller in absolute value than a constant times 1/k, but this is easy to prove, since it's an alternating series with monotonic decreasing absolute value. The error term is smaller in absolute value than the absolute value of the next term, which is smaller than 1/k.) So let's have a close look at pi cos((k+1/2)x) int ------------- dx 1 2 sin(x/2) The denominator doesn't actually change very much, as it grows monotonically from 2*sin(1/2) to 2*sin(pi/2) = 2. The numerator, on the other hand, oscillates rapidly (when k is large) between -1 and +1. So what we'd like to prove is that in each period (from +1 to -1 and back) the integral is very small, so that even when you add up all such periods, the sum is still very small, smaller than a constant times 1/k. So we'll set a = floor((2k+1)/(4*pi)) b = floor((2k+1)/4) (the floor means to round *down* to the nearest integer) and then separate the integral from 1 to pi into many pieces, the first one from 1 to 4(a+1)pi/(2k+1) the next b-a-1 pieces, for m=a+1 to m=b-1, from 4*m*pi/(2k+1) to 4(m+1)pi/(2k+1) the last piece from 4*b*pi/(2k+1) to pi Since (2k+1)/(4*pi) - 1 < a <= (2k+1)/(4*pi) and (2k+1)/4 - 1 < b <= (2k+1)/4 the first and last pieces each have length no bigger than 4pi/(2k+1). Since the integrand cos((k+1/2)x)/(2*sin(x/2)) has absolute value no bigger than 1/(2*sin(1/2)), that means that the integrals on the first and last pieces have absolute value no bigger than 2*pi/((2k+1)sin(1/2)) < pi/sin(1/2) * 1/k. The middle pieces require more attention, since there are many of them, namely b-a-1 of them, and since (2k+1)/(4*pi) - 1 < a b <= (2k+1)/4 that means that there are b-a-1 < (2k+1)/4 - (2k+1)/(4*pi) < (2k+1)/5 of them. So we need to prove that each one is smaller in absolute value than a constant times 1/k^2. They would, in fact, be zero if the denominator were constant, so we should compare them to the integral we would get with a constant denominator. That is, the m'th piece would be 4(m+1)pi/(2k+1) cos((k+1/2)x) int ------------- dx 4*m*pi/(2k+1) 2 sin(x/2) which equals (by the change of variables u = (k+1/2)x ) 2(m+1)pi cos(u) 1 int ------------- ---- du 2*m*pi sin(x/(2k+1)) 2k+1 and since 2(m+1)pi cos(u) 1 int -------------------- ---- du = 0 2*m*pi sin((2m-1)pi/(2k+1)) 2k+1 (this is the constant denominator), that means that our integral (on the m'th piece) equals 2(m+1)pi cos(u) 1 1 int ------ ( ------------- - -------------------- ) du 2*m*pi 2k+1 sin(x/(2k+1)) sin((2m-1)pi/(2k+1)) and now we are going to take absolute values and get a bound on this, because that difference in the integral is very small. In particular, note that |x - (2m-1)pi| <= pi and 1 1 sin v - sin u | ----- - ----- | = | ------------- | sin u sin v (sin u)(sin v) sin((u-v)/2) cos((u+v)/2) = | ------------------------- | (sin u)(sin v) |(u-v)/2| <= -------------- |sin u||sin v| since |sin t| <= |t| for all real numbers t, and |cos t| <= 1. In our case, with u = x/(2k+1) and v = (2m-1)pi/(2k+1), we have |u - v| <= pi/(2k+1) and since 1 <= u <= pi 1 <= v <= pi we also get sin u >= sin(1/2) sin v >= sin(1/2) and therefore 1 1 pi | ----- - ----- | <= ------------------- sin u sin v 2(2k+1)(sin(1/2))^2 and this means that our integral on the m'th piece, which was 2(m+1)pi cos(u) 1 1 int ------ ( ------------- - -------------------- ) du 2*m*pi 2k+1 sin(x/(2k+1)) sin((2m-1)pi/(2k+1)) has absolute value no bigger than 1 pi 2*pi ------ -------------------. 2k+1 2(2k+1)(sin(1/2))^2 Since there are fewer than 2k+1 of these middle pieces (all but the first and last), the sum of all of them can't be bigger than pi^2 ------------------. (2k+1)(sin(1/2))^2 And since the first and last pieces are no bigger than 2*pi/((2k+1)sin(1/2)) that means that the sum of all of them, which is pi cos((k+1/2)x) int ------------- dx 1 2 sin(x/2) is no bigger than 1 pi^2 2*pi ------ ( ------------ + -------- ) (2k+1) (sin(1/2))^2 sin(1/2) which is less than 1 pi^2 pi - ( ------------- + -------- ) k 2(sin(1/2))^2 sin(1/2) thus proving that the integral approaches zero as k grows to infinity, as we were trying to do. And that does it! So we have proven that the sum converges and equals -ln(2) - ln(sin(1/2)) = 0.042019505825368961725798384.... If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
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