# Proof of faulhaber's formula

1. May 6, 2010

### the one

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

how is $$\left(\sum_{k=0}^{\infty}\frac{n^{k+1}}{k+1}.\frac{x^k}{k!}\right)\left(\sum_{l=0}^{\infty}B_{l}\frac{x^l}{l!}\right)$$ equals $$\sum_{k=0}^{\infty}\left(\sum_{i=0}^{k}\frac{1}{k-i+1}\binom{k}{i}B_{i}n^{k+1-i}\right)\frac{x^k}{k!}$$
??

Last edited by a moderator: May 4, 2017
2. May 6, 2010

### lurflurf

Effect the change of variable
i=l
k'=k+l

3. May 6, 2010

### Martin Rattigan

I don't know how that would work with the limits. Wouldn't you still have $\sum_{i=0}^\infty$?

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

$(\sum_{k=0}^\infty a_kx^k)(\sum_{l=0}^\infty b_lx^l)=\sum_{k=0}^\infty(\sum_{i=0}^k a_{k-i}b_i)x^k$

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

Replacing the $a_i$ and $b_i$ from the given sequences and using

$\left( \begin{array}{c} k\\ i \end{array} \right )=\frac{k!}{(k-i)!i!}$

gives you what you want.

Last edited: May 6, 2010