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Proof of faulhaber's formula

  1. May 6, 2010 #1
    hi
    i have a problem with the the proof of faulhaber's formula given http://planetmath.org/encyclopedia/ProofOfFaulhabersFormula.html" [Broken]

    how is [tex]\left(\sum_{k=0}^{\infty}\frac{n^{k+1}}{k+1}.\frac{x^k}{k!}\right)\left(\sum_{l=0}^{\infty}B_{l}\frac{x^l}{l!}\right)[/tex] equals [tex]\sum_{k=0}^{\infty}\left(\sum_{i=0}^{k}\frac{1}{k-i+1}\binom{k}{i}B_{i}n^{k+1-i}\right)\frac{x^k}{k!}[/tex]
    ??
    thanks in advance :)
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 6, 2010 #2

    lurflurf

    User Avatar
    Homework Helper

    Effect the change of variable
    i=l
    k'=k+l
     
  4. May 6, 2010 #3
    I don't know how that would work with the limits. Wouldn't you still have [itex]\sum_{i=0}^\infty[/itex]?

    A simpler? if longer explanation is that the sequence on the right is the Cauchy product of the two on the left:

    [itex](\sum_{k=0}^\infty a_kx^k)(\sum_{l=0}^\infty b_lx^l)=\sum_{k=0}^\infty(\sum_{i=0}^k a_{k-i}b_i)x^k[/itex]

    which is taking the coefficient of [itex]x^k[/itex] on the RHS as the sum of all the [itex]a_{k-i}b_i[/itex] with [itex]0\leq i\leq k[/itex]. This converges if both series on the left do and at least one (in this case both) are absolutely convergent.

    Replacing the [itex]a_i[/itex] and [itex]b_i[/itex] from the given sequences and using

    [itex]\left( \begin{array}{c} k\\ i \end{array} \right )=\frac{k!}{(k-i)!i!}[/itex]

    gives you what you want.
     
    Last edited: May 6, 2010
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