suppose we define the Bernoulli numbers $b_n, n = 1, 2, 3, \ldots$ by the Faulhaber's fomula
$$\begin{eqnarray*}1 + 2^k + 3^k + \ldots n^k &=& \frac{1}{k+1}[n^{k+1} + b_1 c(k+1,2) n^k + b_2 c(k+1,3)n^{k-1} + \ldots ]\\&=& \frac{(n+b)^{k+1}-b^{k+1}}{k+1}\end{eqnarray*}$$
with the proviso we replace $b^k$ by $b_k$ in the binomial expansion and $c(k,l)$ is $k$ choose $l.$
my question is, how do you show all the odd Bernoulli numbers except $b_1$ is zero without invoking heavy analytical tools so that the reason can be explained to
a student who has only had one or two semesters of calculus in high school.
Best Answer
Why not take the opportunity to expose them to the glory of generating functions?
Define the Bernouli numbers to be the coefficients in $$\frac{t}{1-e^{-t}} = \sum B_m \frac{t^m}{m!}.$$
Prove that these give the sum of the $k$-th powers by equating coefficients of $x^k$ in $$\sum_{j=1}^n \frac{j^k x^k}{k!} = \sum_{j=1}^n e^{jx} = \frac{e^{(n+1)x} - e^x}{e^x-1}$$ $$= \left( \frac{e^{nx} -1}{x} \right) \left( \frac{x}{1-e^{-x}} \right) = \sum_{\ell=0}^{\infty} \frac{n^{\ell+1} x^{\ell}}{(\ell+1)!} \cdot \sum B_m \frac{t^m}{m!}.$$
Then prove that the odd Bernoulli numbers vanish by noting that $$\sum B_m \frac{t^m}{m!} - \sum B_m \frac{(-t)^m}{m!} = \frac{t}{1-e^{-t}} - \frac{-t}{1-e^t} = \frac{t (e^t-1)}{e^t-1} = t.$$