anonymous
  • anonymous
find a formula for the nth parial sum of the series and use it to find the series' sum if convergent: (1/2*3) + (1/3*4) + (1/4*5) +...+ {1/(n+1)(n+2)}+... and then could you tell me how you did it =)?
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
Well, the first part is easy, a formula is you write the Riemann Sum symbol which looks like a capital E. You put n on top of the E and k=1 at the bottom of the E. Next to the E your 1/[(k+1)(k+2)]=(1/2*3) + (1/3*4) + (1/4*5) +...+ {1/(n+1)(n+2)}
anonymous
  • anonymous
awesome, thanks a lot. that gets me started
anonymous
  • anonymous
slight correction (telescoping series) k=0, by partial fractions we rewrite that Reimann sum [1/(k+1) -1/(k+2)} Nth partial sum=(1-(1/2) =(1/2 - 1/3) + ... +[(n+1)^-1 - (n+2)^-1]= 1 - (n+2)^-1

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anonymous
  • anonymous
lim as n goes infinity of 1 - 1/(n+2)=1. The series converges to 1

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