Nasty sum

  • Thread starter bogdan
  • Start date
  • #1
bogdan
191
0
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 ?
 

Answers and Replies

  • #2
plus
178
1
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.
 
  • #3
rgrig
34
1
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
 
  • #4
bogdan
191
0
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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