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

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SUMMARY

The summation of sin(x/[n*(n+1)]) over n from 1 to infinity simplifies to tan(x). As n increases, each term in the sum approaches x/((n*n+1)), which can be approximated using Taylor expansion. The transformation sin(x/(n(n+1))) = sin(x/n)cos(x/(n+1)) - sin(x/(n+1))cos(x/n) leads to the conclusion that the series converges to tan(x) through a series of cancellations. This result is confirmed through the analysis of the series behavior as n approaches infinity.

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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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