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Nice sum (Posted on 2004-08-24) Difficulty: 3 of 5
Give a closed expression for the infinite sum:

1/1 + 2/(2+3) + 3/(4+5+6) + 4/(7+8+9+10) + ...

  Submitted by Federico Kereki    
Rating: 2.7500 (4 votes)
Solution: (Hide)
The nth term is n/(k+1+k+2+...+k+n) where k=n(n-1)/2; this works out to n/(nk+n(n+1)/2)= 1/(k+(n+1)/2)= 2/(n²+1).

So, we need to calculate the sum for n>0 of 2/(n²+1), which is the same as -1+ the sum for n=-∞ to +∞ of 1/(n²+1).

The latter sum equals (proof) πcoth(π), so the final result is πcoth(π)-1.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionProblem Solution With ExplanationK Sengupta2007-05-28 12:15:57
answerK Sengupta2007-05-28 12:14:40
re: SolutionSarifa2004-10-08 17:48:13
Some Thoughtsre: SolutionNick Hobson2004-09-01 14:51:37
SolutionKeith Steele2004-08-30 20:10:09
re: SolutionFederico Kereki2004-08-28 21:24:33
SolutionSolutionNick Hobson2004-08-27 16:34:34
solutionbernie2004-08-26 13:53:49
Questionre: starting observationVee-Liem Veefessional2004-08-26 00:27:58
Questionre(2): No SubjectVee-Liem Veefessional2004-08-26 00:23:03
re: No SubjectFederico Kereki2004-08-25 08:51:47
QuestionNo SubjectVee-Liem Veefessional2004-08-24 23:36:59
Some Thoughtsstarting observationCharlie2004-08-24 12:54:58
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