## #math - Sun 15 Apr 2007 between 00:03 and 00:42

### NY Lost Funds

 lesshaste well ,,, I changed the lines to 3 320 120 55.92 (i.e. I removed the names)and I did scan(file = "scores.txt")but it doesn't seem to have loaded it in in a useful wayanyone able to help? TRWBW lesshaste: this isn't really the channel for that. it's more for the theory than how to use a particular program. try something like #stats i'm guessing. lesshaste sadly no such channel TRWBW about as likely to get your answer here as in #klingon-speakers lesshaste ohR is quite popular amongst mathematicians WILDSTYLE R is alright. lesshaste can I ask a question about the coupon collector's problem? WILDSTYLE I'm kinda partial to C. lesshaste WILDSTYLE: is that a joke? TRWBW i tried R, but when back to using C and Fortran JabberWalkie lesshaste: noyou have to read the topic first TRWBW lesshaste: no, for statistics it's a lot nicer to do it in a real programming language. it runs faster, you can get libraries with better algorithms, etc. Zanco-afIRC Hello Everybody WILDSTYLE hello. jadenbane Does Sum (1, inf., 1/(n(n+1))) = 0 ? Mulder jadenbane,nofor starters, 1/2 is one of the terms you addso automatically, your sum is > 0 jadenbane Haha, I should actually look at the numbers shouldn't I :PSum (1, inf., 1/(n(n+1))) = Sum (1, inf., 1/n - 1/(n+1)) which makes me think there will be cancelation. Mulder if you can decompose it into a partial fraction lik ethatyou can telescope, and work out what the infinite sum isand and your final sum is probably something like 1 jadenbane See, I'm not sure what telescoping is, since I only have a vague notion of ir.*it. Mulder well, a series is just a seqeuence of partial fractions jadenbane Oh, that sum will give 1/1 - 1/2 + 1/2 + 1/3 - 1/3 etc. Mulder yes jadenbane So yeah, I was hasty in guessing zero. I should have guessed one. Mulder so s_n = 1 - 1/2 + 1/2 + 1/3 - 1/3 + ... - 1/(n-1) + 1/(n-1) - 1/ns_n = 1 + 1/nthen if we take the limit of that, we get 1since 1/n goes to 0 as n -> Infinity