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

[bogdan]

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 ?

 

[plus]

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.

 

[rgrig]

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

 

[bogdan]

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)