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Summation by Parts

  1. Jun 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Using summation by parts, find Sum[n/3^n].


    2. Relevant equations

    Sum[a_k*b_k] = s_n*b_(n+1) - Sum[s_k(b_(k+1)-b_k]


    3. The attempt at a solution

    Let a_k = 1/3^k and b_k = k. Then b_(k+1)-b_k = 1. But what is s_k? I know that it is 1/3 + 1/3^2 + 1/3^3 + ... but what is the general term? Thanks for your help.
     
  2. jcsd
  3. Jun 24, 2010 #2
  4. Jun 25, 2010 #3
    S_n is the sequence of partial sums of a_k. My formula is from Goldberg's Methods of Real Analysis.
     
  5. Jun 25, 2010 #4
    sum(n*(1/3)^n)
    very similar to a geometric series, after one differentiation.

    sum(n*(1/3)^n) = (1/3)*sum(n*(1/3)^(n-1))

    we know that the sum of a geometric series is 1/(1-q), here q=1/3.

    sum = (1/3)*diff(1/(1-(1/3)))

    sound familiar?

    *it's sum found using the integration/differentiation by parts theorem.
     
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