Summation of sin(x/[n*(n+1)]) over n from 1 to ∞

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Discussion Overview

The discussion revolves around the summation of the expression sin{x/[n*(n+1)]}/[cos(x/n)*cos(x/(n+1))] as n approaches infinity, with x being a constant. Participants explore potential solutions and approximations related to this mathematical series.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Technical explanation

Main Points Raised

  • One participant proposes the sum of sin{x/[n*(n+1)]}/[cos(x/n)*cos(x/(n+1))] and seeks ideas for evaluation.
  • Another participant suggests that as n becomes large, each term in the sum approximates to (x/((n*n+1))) using Taylor expansion, and discusses the potential for truncation error in summing to a certain point.
  • A different participant claims that the result of the summation is tan(x), providing a derivation involving sin and cos identities that leads to a telescoping series of tangent terms.
  • One participant agrees with the conclusion that the result is tan(x) and mentions curiosity about the problem's origin from an analysis exercise book.

Areas of Agreement / Disagreement

There appears to be agreement among some participants that the result of the summation is tan(x), while the initial proposal and approximation methods suggest a more exploratory approach. The discussion includes varying levels of certainty regarding the solution.

Contextual Notes

Some assumptions regarding the convergence of the series and the validity of the Taylor expansion are not explicitly stated, which may affect the interpretation of the results.

bogdan
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sum sin{x/[n*(n+1)]}/[cos(x/n)*cos(x/(n+1))], where n goes from1 to infinity and x is a given constant...
Any ideas ?
 
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Well clearly as n gets large, each element in the sum tends to (x/((n*n+1))) using taylor expansion.

The sum from k to infinity of (x/(n*(n+1))) is x/k, so you could sum up to a certain point and then use this to approximate the truncation error.
 
Hello bogdan,

I think the answer is tan(x).

sin(x/(n(n+1))) = sin(x/n)cos(x/(n+1)) - sin(x/(n+1))cos(x/n)

After some simplifications you get:

sum tan(x/n)-tan(x/(n+1))

That is:

tan x - tan x/2 +
+ tan x/2 - tan x/3
...

which is

tan x - tan 0 = tan x
 
Yes...it's tan(x)...
(eram doar curios sa vad cine stie sa-l rezolve...e dintr-o carte de exercitii de analiza...cu tot cu raspunsuri)
 

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