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anonymous

  • 5 years ago

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 =)?

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  1. anonymous
    • 5 years ago
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    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)}

  2. anonymous
    • 5 years ago
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    awesome, thanks a lot. that gets me started

  3. anonymous
    • 5 years ago
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    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

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

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