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

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The discussion revolves around the evaluation of the sum sin{x/[n*(n+1)]}/[cos(x/n)*cos(x/(n+1))] as n approaches infinity, with x as a constant. Participants note that as n increases, each term in the sum approximates to x/((n*n+1)), which can be analyzed using Taylor expansion. The sum converges to a form that suggests a relationship with the tangent function. Through simplification, it is concluded that the overall result of the sum is tan(x). The conversation highlights the mathematical manipulations leading to this conclusion, emphasizing the connection between the series and the tangent function.
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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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