# proof of Faulhaber’s formula

###### Theorem 0.1.

If $k\in\mathbb{N},2\leq n\in\mathbb{Z}$, then

 $\sum_{m=1}^{n-1}m^{k}=\frac{1}{k+1}\sum_{i=0}^{k}\binom{k+1}{i}B_{i}n^{k+1-i}=% \int_{1}^{n}b_{k}(x)dx$

where the $B_{i}$ are the Bernoulli numbers and $b_{i}$ the Bernoulli polynomials.

The exponential generating function for the Bernoulli numbers is

 $\sum_{n=0}^{\infty}B_{n}\frac{x^{n}}{n!}=\frac{x}{e^{x}-1}$

We develop an equation involving sums of Bernoulli numbers on one side, and a simple generating involving powers of $e$ that gives us the appropriate sum of powers on the other side. Equating coefficients of powers of $x$ then gives the result.

To get a generating function where the coefficient of $x^{n}/n!$ is $\sum_{m=1}^{n-1}m^{k}$, we can use

 $\displaystyle\sum_{m=0}^{n-1}e^{mx}$ $\displaystyle=\sum_{m=0}^{n-1}\sum_{k=0}^{\infty}\frac{m^{k}x^{k}}{k!}$ $\displaystyle=\sum_{k=0}^{\infty}\left(\sum_{m=1}^{n-1}m^{k}\right)\frac{x^{k}% }{k!}$

But this is also a geometric series, so

 $\displaystyle\sum_{k=0}^{n-1}e^{kx}$ $\displaystyle=\frac{1-e^{nx}}{1-e^{x}}$ $\displaystyle=\frac{e^{nx}-1}{x}\cdot\frac{x}{e^{x}-1}$ $\displaystyle=\frac{e^{nx}-1}{x}\sum_{l=0}^{\infty}B_{l}\frac{x^{l}}{l!}$ $\displaystyle=\left(\sum_{k=0}^{\infty}\frac{n^{k+1}}{k+1}\cdot\frac{x^{k}}{k!% }\right)\left(\sum_{l=0}^{\infty}B_{l}\frac{x^{l}}{l!}\right)$ $\displaystyle=\sum_{k=0}^{\infty}\left(\sum_{i=0}^{j}\frac{1}{k-i+1}\binom{k}{% i}B_{i}n^{k+1-i}\right)\frac{x^{k}}{k!}$

Equating coefficients of $x^{k}/k!$ we get

 $\displaystyle\sum_{m=1}^{n-1}m^{k}$ $\displaystyle=\sum_{i=0}^{k}\frac{1}{k-i+1}\binom{k}{i}B_{i}n^{k+1-i}$ $\displaystyle=\sum_{i=0}^{k}\frac{k!}{(k-i+1)i!(k-i)!}B_{i}n^{k+1-i}=\frac{1}{% k+1}\sum_{i=0}^{k}\binom{k+1}{i}B_{i}n^{k+1-i}$

which proves the first equality.

If $f(x)$ is a polynomial, write $[x^{r}]f(x)$ for the coefficient of $x^{r}$ in $f(x)$. Then

 $[x^{r}]b_{k}(x)=\frac{1}{r}[x^{r-1}]b_{k}^{\prime}(x)=\frac{k}{r}[x^{r-1}]b_{k% -1}(x)$

and thus if $r\leq k$, iterating, we get

 $[x^{r}]b_{k}(x)=\binom{k}{r}[x^{0}]b_{k-r}(x)=\binom{k}{r}B_{k-r}$

Then using the fact that $b_{k}^{\prime}=kb_{k-1}$, we have

 $\displaystyle\int_{1}^{n}b_{k}(x)$ $\displaystyle=\frac{1}{k+1}(b_{k+1}(n)-b_{k+1}(1))=\frac{1}{k+1}\sum_{r=0}^{k+% 1}[x^{r}]b_{k+1}(x)(n^{r}-1)$ $\displaystyle=\frac{1}{k+1}\sum_{r=0}^{k+1}\binom{k+1}{r}B_{k+1-r}(n^{r}-1)=% \frac{1}{k+1}\sum_{r=1}^{k+1}\binom{k+1}{r}B_{k+1-r}n^{r}$

Now reverse the order of summation (i.e. replace $r$ by $k+1-r$) to get

 $\int_{1}^{n}b_{k}(x)=\frac{1}{k+1}\sum_{r=0}^{k}\binom{k+1}{k+1-r}B_{r}n^{k+1-% r}=\frac{1}{k+1}\sum_{r=0}^{k}\binom{k+1}{r}B_{r}n^{k+1-r}$
Title proof of Faulhaber’s formula ProofOfFaulhabersFormula 2013-03-22 18:43:50 2013-03-22 18:43:50 rm50 (10146) rm50 (10146) 4 rm50 (10146) Theorem msc 11B68