From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: looking for series that sums to one Date: 3 Feb 2001 21:40:22 -0500 Newsgroups: sci.math Summary: Slowly decreasing positive series which sum to 1 In article <95iblu$12b$1@nnrp1.deja.com>, wrote: :Mathematics is a fond memory fading fast, so :please excuse me if this question is too simple :or not well expressed. I am looking for a simple :series that satisfies the following conditions. : :1) Series sums to one. :2) The terms are positive. :3) The terms decrease slowly. :4) The terms decrease "evenly". : :My first try was : : 1/2 + 1/4 + 1/8 + ... : :but this decreases too quickly. My second try was : : 1/4 + 1/4 + 1/8 + 1/8 + 1/16 + 1/16 + ... : :and its direct alternatives like starting with 4 :1/8's or 8 1/16's, but this series doesn't :decrease "evenly". My third try was : : 2^(-1/k) + 2^(-2/k) + 2^(-3/k) + ... : :with k set to slow the rate of decrease to :taste. This deries satisfies the last three :conditions, I think, but now doesn't sum to one. Another favorite from textbooks is 1/(1*2) + 1/(2*3) + ... + 1/(n*(n+1)) + ... it is slower than any convergent geometric series. The secret is that 1/(1*2) = 1/1 - 1/2, 1/(2*3) = 1/2 - 1/3, 1/(3*4) = 1/3 - 1/4, and so on. The terms go to 0 very "evenly" : the repeated differences are all monotone. If you want something slower, at the price of not being rational, here is (1/sqrt(1) - 1/sqrt(2)) + (1/sqrt(2) - 1/sqrt(3)) + ... with general term (1/sqrt(n) - 1/sqrt(n+1)) , and obvious variants to slow it down even more: try general term, from n=2, looking like (log(2)/log(n) - log(2)/log(n+1)). Have fun, ZVK(Slavek). ============================================================================== From: Jan Kristian Haugland Subject: Re: looking for series that sums to one Date: Sun, 04 Feb 2001 19:42:19 +0100 Newsgroups: sci.math Bill Taylor wrote: > vanegas@my-deja.com writes: > |> 1) Series sums to one. > |> 2) The terms are positive. > |> 3) The terms decrease slowly. > |> 4) The terms decrease "evenly". > > They don't come all that much slower than this very "even" series... > > 1 = 1/(1.2) + 1/(2.3) + 1/(3.4) + 1/(4.5) + ... Not _much_ slower, but a _little_ slower: 1 = 1/(2*3/2) + 1/(3*4/2) + 1/(4*5/2) + 1/(5*6/2) + ... -- Jan Kristian Haugland http://home.hia.no/~jkhaug00 ============================================================================== From: ol3@webtv.net (Oscar Lanzi III) Subject: Re: looking for series that sums to one Date: Sun, 4 Feb 2001 19:33:02 -0600 (CST) Newsgroups: sci.math sum[n=1 to inf] [1/(sqrt(n)*sqrt(n+1)*(sqrt(n)+sqrt(n+1))] = 1 Why? Because a_n = 1/sqrt(n) - 1/sqrt(n+1). The ingenuity in proposing solutions is in disguising the difference. Terms go as n^(-3/2) which beats the 1/(n*(n+1)) series for slow conergence. --OL ============================================================================== From: "David Petry" Subject: Re: looking for series that sums to one Date: Sun, 4 Feb 2001 14:14:04 -0800 Newsgroups: sci.math >> vanegas@my-deja.com writes: >> |> 1) Series sums to one. >> |> 2) The terms are positive. >> |> 3) The terms decrease slowly. >> |> 4) The terms decrease "evenly". Not sure what "evenly" means, but here's the answer that I'm sure Leroy Quet would love. Let H_k = 1 + 1/2 + 1/3 + ... + 1/k and let T_n = 1/( (n+1) * T_n * T_{n+1} ) then {T_n} is a sequence of positive slowly decreasing terms and T_1 + T_2 + T_3 ... = 1 ============================================================================== From: "David Petry" Subject: Re: looking for series that sums to one Date: Mon, 5 Feb 2001 21:41:48 -0800 Newsgroups: sci.math Bill Taylor wrote in >"David Petry" writes: > >|> here's the answer that I'm sure Leroy Quet would love. > >Yes! Beat him to the punch this time! > > >|> Let H_k = 1 + 1/2 + 1/3 + ... + 1/k >|> and let T_n = 1/( (n+1) * T_n * T_{n+1} ) Wow! Did I write that? (Yes, I did) Clearly I meant and let T_n = 1/( (n+1) * H_n * H_{n+1} ) >|> and T_1 + T_2 + T_3 ... = 1 > >That is waaaay kool! Full marks David! Thanks for the award, but what I really want is a chocolate fish. > > > _ /\./|_ > _.-| |\ | / |_ > / \ _>-"""-._.'|_ > |`-.' `./ \ > /`./ \-<. > \_| Sci.math |_/ > /_| |_\ > ) | Special | | > \'_\ Award /`< > |_/`. .'\_/ > \_/ >-.._..-'\_| > `-\_| \_\|_/ > | : \/.| .| > | : .| .| > | : .| .| > | : .| .| > | : .| .| > | :/\ .| .| > | /| \.|\.| > |/ |/ \| \| > >------------------------------------------------------------------------------ > Bill Taylor W.Taylor@math.canterbury.ac.nz >------------------------------------------------------------------------------ > Stamp out silly signatures ! >------------------------------------------------------------------------------ ============================================================================== From: "David Petry" Subject: Re: looking for series that sums to one Date: Tue, 6 Feb 2001 21:01:10 -0800 Newsgroups: sci.math Bill Taylor wrote in >"David Petry" writes: > >|> Wow! Did I write that? (Yes, I did) Clearly I meant >|> and let T_n = 1/( (n+1) * H_n * H_{n+1} ) > >Yes I spotted that, but thought it would be churlish to point it out >during an award ceremony. > >BTW: If you can get one like 1/(n.logn.(loglogn)^2) that would ROCK! Hmmm, you haven't figured out how this game is played yet? Let H_k = 1/1 + 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ... + 1/p_{k-1} where p_k is the k'th prime. Let T_n = 1/( p_n * H_n * H_{n+1} ) Then, assuming I haven't done anything stupid, T_1 + T_2 + T_3 ... = 1 and T_k decreases as fast as you asked for. ============================================================================== From: David Petry Subject: Re: Those ultra-slowly diverging series. Date: Mon, 08 Oct 2001 17:02:43 -0700 Newsgroups: sci.math Bill Taylor wrote: > > David Petry writes: > |> Bill Taylor wrote: > |> > > |> > From time to time a series is mentioned here that converges even more > |> > slowly than all those of the type > |> > t3 = 1/[n(log n)(log log n)(log log log n)] . > |> > > [namely] > |> > 1 > |> > v = ------------------------------------------------------ [2] > |> > n n(log n)(log log n)(log log log n)....(log log..[>1]..log n) > |> > > > |> Note that the term that is squared must be less than e, and from there it > |> is easily seen that the series diverges, assuming the first series you > |> give does in fact diverge. > > Oh of course how obvious! Thanks. > > So given that the various t-series above all DIverge but very slowly, > and given that their adjustments by powering the last multiplicands > CONverge, similarly slowly, and given that all the "diagonal" series > like v above also DIverge; can we find an explicit series with an explicit > formula like v has, that CONverges slower than all the adjusted-t's? > > I realize that they must exist by diagonalization, I was just hoping for > an explicit formulaic example. Define K(x) = 0 if x < 0, else K(x) = 1 + K(log(x)). Loosely, K(x) is the number of iterations of the log function needed to reduce x to 0. Then sum v_n/K(n) diverges, but sum v_n/K(n)^2 converges, and it converges more slowly than any of the "adjusted t series" you mentioned. [slightly reformatted -- djr