proof of Faulhaber’s formula

Theorem 0.1.

If $k\mathrm{\in }\mathrm{N}\mathrm{,}\mathrm{2}\mathrm{\le }n\mathrm{\in }\mathrm{Z}$, then

$$\sum _{m=1}^{n-1}{m}^k=\frac{1}{k+1}\sum _{i=0}^k\left(\genfrac{}{}{0pt}{}{k+1}{i}\right)B_i{n}^{k+1-i}={\int }_1^n{b}_k(x)𝑑x$$

where the $B_i$ are the Bernoulli numbers and $b_i$ the Bernoulli polynomials.

The exponential generating function for the Bernoulli numbers is

$$\sum _{n=0}^{\mathrm{\infty }}{B}_n\frac{x^n}{n!}=\frac{x}{e^x-1}$$

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}}{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{\displaystyle \frac{m^k{x}^k}{k!}}$

$={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\left({\displaystyle \sum _{m=1}^{n-1}}{m}^k\right){\displaystyle \frac{x^k}{k!}}$

But this is also a geometric series, so

${\displaystyle \sum _{k=0}^{n-1}}{e}^{kx}$$={\displaystyle \frac{1-e^{nx}}{1-e^x}}$

$={\displaystyle \frac{e^{nx}-1}{x}}\cdot {\displaystyle \frac{x}{e^x-1}}$

$={\displaystyle \frac{e^{nx}-1}{x}}{\displaystyle \sum _{l=0}^{\mathrm{\infty }}}{B}_l{\displaystyle \frac{x^l}{l!}}$

$=\left({\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{\displaystyle \frac{n^{k+1}}{k+1}}\cdot {\displaystyle \frac{x^k}{k!}}\right)\left({\displaystyle \sum _{l=0}^{\mathrm{\infty }}}{B}_l{\displaystyle \frac{x^l}{l!}}\right)$

$={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\left({\displaystyle \sum _{i=0}^j}{\displaystyle \frac{1}{k-i+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{i}}\right)B_i{n}^{k+1-i}\right){\displaystyle \frac{x^k}{k!}}$

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

${\displaystyle \sum _{m=1}^{n-1}}{m}^k$$={\displaystyle \sum _{i=0}^k}{\displaystyle \frac{1}{k-i+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{i}}\right)B_i{n}^{k+1-i}$

$={\displaystyle \sum _{i=0}^k}{\displaystyle \frac{k!}{(k-i+1)i!(k-i)!}}{B}_i{n}^{k+1-i}={\displaystyle \frac{1}{k+1}}{\displaystyle \sum _{i=0}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k+1}{i}}\right)B_i{n}^{k+1-i}$

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 $r\le k$, iterating, we get

$$[x^r]b_k(x)=\left(\genfrac{}{}{0pt}{}{k}{r}\right)[x^0]b_{k-r}(x)=\left(\genfrac{}{}{0pt}{}{k}{r}\right)B_{k-r}$$

Then using the fact that $b_k^{\prime }=k{b}_{k-1}$, we have

${\displaystyle {\int }_1^n}{b}_k(x)$$={\displaystyle \frac{1}{k+1}}(b_{k+1}(n)-b_{k+1}(1))={\displaystyle \frac{1}{k+1}}{\displaystyle \sum _{r=0}^{k+1}}[x^r]b_{k+1}(x)(n^r-1)$

$={\displaystyle \frac{1}{k+1}}{\displaystyle \sum _{r=0}^{k+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{k+1}{r}}\right)B_{k+1-r}(n^r-1)={\displaystyle \frac{1}{k+1}}{\displaystyle \sum _{r=1}^{k+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{k+1}{r}}\right)B_{k+1-r}{n}^r$

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

$${\int }_1^n{b}_k(x)=\frac{1}{k+1}\sum _{r=0}^k\left(\genfrac{}{}{0pt}{}{k+1}{k+1-r}\right)B_r{n}^{k+1-r}=\frac{1}{k+1}\sum _{r=0}^k\left(\genfrac{}{}{0pt}{}{k+1}{r}\right)B_r{n}^{k+1-r}$$