From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: convergent series? Date: 15 Jan 1999 18:34:23 GMT Keywords: Simple tests for series convergence in a tough case Just in case any calculus students were following this, I'd like to correct what I think is being claimed here. The original question was >How can be proved (or disproved) that the series > > inf > S (-1)^n abs(sin(n))/n converge? > n=1 I don't know whether it converges or not. Evidently someone suggested using a comparison test, for Tom Hardy wrote: >Perhaps my explanation was unclear. The resulting series is -(1 - 1/2 + >1/3 -1/4 +... ) which converges to -ln(2). This illustrates the series >converges, but as you say not absolutely, since then it would be the >harmonic series to which you referred. This is invalid reasoning on two counts. We can attempt to test for convergence using the inequality | sin(x) | <= 1. Well, fine: the original series converges absolutely iff the series Sum( |sin(n)|/n ) converges. That series, in turn, would converge if Sum( 1 / n ) converged, since |sin(n)|<=1. But the latter series diverges. Thus we have no conclusion. One could imagine that a_n = |sin(n)| came out to be, oh, about as big as 1/n in general (that's not true BTW, but it illustrates the reasoning); in that case Sum( a_n/n ) would converge even though Sum( 1/n ) diverged. So we have no easy way to test for absolute convergence. Further, it appears that the suggestion is being made that a comparison test can be used for series with mixed signs. This is false too. Consider the series Sum( (-1)^n a_n ) where a_n is, say, 1/n^2 when n is odd, but 1/n when n is even. In this case, each a_n is (positive and) no bigger than 1/n. Since the series Sum( (-1)^n / n ) converges (conditionally, to log(2) ), one might think a comparison test would imply that the other sequence Sum( (-1)^n a_n ) also converged. One would be wrong. The negative terms in this sum converge but the positive ones do not, so the whole series diverges to +infinity. So we have no easy way to test for conditional convergence, either. I can't imagine a reasonable test for convergence which would cover terms involving (-1)^n |sin(n)|. Fortunately, I can't imagine a natural problem which would call for this, either. dave ============================================================================== From: "Kevin Lacker" Newsgroups: sci.math Subject: Re: convergent series? Date: Fri, 15 Jan 1999 17:12:38 -0500 It's not absolutely convergent. Just consider the absolute values of sin n. min(|sin n|,|sin (n+1)|) will be bounded below, say by m > 0, and then it's pretty clear that the absolute values diverge. It is conditionally convergent, however; use Abel's technique (I think) of showing the partial sums of ((-1)^n)(sin n) are bounded and the 1/n approach zero. The first part you can just use some trig identities for, I'm sure. Dave Rusin wrote in message <77o1nf$pg0$1@gannett.math.niu.edu>... >Just in case any calculus students were following this, I'd like to >correct what I think is being claimed here. > >The original question was >>How can be proved (or disproved) that the series >> >> inf >> S (-1)^n abs(sin(n))/n converge? >> n=1 > >I don't know whether it converges or not. Evidently someone suggested >using a comparison test, for Tom Hardy wrote: ============================================================================== From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Newsgroups: sci.math Subject: Re: convergent series? Date: 15 Jan 1999 16:27:14 -0500 In article <778b4t$o8v$1@winter.news.rcn.net>, [Tom Hardy's contribution deleted] The original article has no longer a pointer to it here, so I branch off, with apologies. >Stéphane Ménart wrote in message <36966F29.9FD@club-internet.fr>... >>How can be proved (or disproved) that the series >> >> inf >> S (-1)^n abs(sin(n))/n converge? >> n=1 >> >>(where abs() denotes the absolute value) [...] This series exhibits rather irregular behaviour. It takes a trick to show that the series does not converge absolutely, that is, that sum[n=1 to inf.] abs(sin(n))/n diverges. To show that, write abs(sin(n))/n = a(n) + b(n) where a(n) = (abs(sin(n)) - abs(sin(n-1))) / (2*n) b(n) = (abs(sin(n)) + abs(sin(n-1))) / (2*n) Now, using summation by parts, we can show that the sum of a(n) is convergent, while the sum of b(n) diverges to infinity. The main ingredient in proving the divergence is that h: h(x) = abs(sin(x)) + abs(sin(x-1)) is a function bounded below by sin(1) which is positive. The original question is harder, though. The two conditional convergence tests I know are not powerful enough (unless someone invents another trick). Dirichlet's Test: Let u(n) be terms of a sequence convergent monotonically to 0, and let v(n) be terms of a sequence whose partial sums (v(1) + ... + v(n)) are bounded by the same constant. Then the sum of u(n)*v(n) is convergent. Abel's Test: Let u(n) be terms of a convergent series, and let v(n) be terms of a monotone convergent sequence. Then the sum of u(n)*v(n) is convergent. (Dirichlet's test is powerful enough to show that the sum of sin(n)/n is convergent, and that the sum of a(n) above is convergent.) Good luck, ZVK(Slavek).