How does the Cauchy product prove Faulhaber's formula?

[the one]

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

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!}
??
thanks in advance :)

 

[lurflurf]

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

 

[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.