From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: Sum_{n=1}^{\infty}(cos(nx)/n) Date: 20 Dec 2000 01:19:20 -0500 Newsgroups: sci.math Summary: [missing] In article , Wade Ramey wrote: :In article , :Ken.Pledger@vuw.ac.nz (Ken Pledger) wrote: : :> Then f'(x) is a :> geometric series. : :Which fails to converge for all real x. But you got the right :answer, you Euler you. : :Let's throw in a little rigor: For complex z, |z| < 1, : : 1 + z + z^2 + ... + z^n + .... = 1/(1 - z). : :We can integrate term by term to get : : z + z^2/2 + ... + z^n/n + .... = -log (1 - z) : :for |z| < 1, where log denotes the principal branch. So replacing :z by exp(ix), x in [-pi, pi]\{0}, will give Ken's answer, assuming :the series still converges to the desired limit. What's the :easiest way to see that this happens gang? : :Wade Good old summation by parts: replace cos(n*x) by (sin((n+1/2)*x) - sin((n-1/2)*x)) / (2*sin(x/2)) wherever the latter is defined. The trick is: transform the sum from 1 to N and observe that the right "boundary term" goes to 0 as N goes to infinity. Result (if my calculations are correct): -1/2 + sum[n=1 to infinity] sin((n+1/2)x)/(2*n*(n+1)*sin(x/2)) Except for x in {integer multiples of 2*pi}, the series converges absolutely, and also locally uniformly (but slowly). Reformulated: after the series is multiplied by sin(x/2), it will converge absolutely and uniformly everywhere on the real line. Cheers, ZVK(Slavek)