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Nasty sum

  1. Apr 30, 2003 #1
    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 ?
     
  2. jcsd
  3. Apr 30, 2003 #2
    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.
     
  4. May 2, 2003 #3
    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
     
  5. May 2, 2003 #4
    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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