Sum of Bernoulli Numbers by Binomial Coefficients Vanishes

Theorem

 * $\displaystyle \sum_{k \mathop = 0}^{n - 1} \binom n k B_k = 0$

where $B_k$ denotes the $k$th Bernoulli number.

Proof
Take the definition of Bernoulli numbers:
 * $\displaystyle \frac x {e^x - 1} = \sum_{n \mathop = 0}^\infty \frac{B_n x^n} {n!}$

From the definition of the exponential function:

Thus:

For $n > 1$, the coefficients for $x^{n - 1}$ in the product on the equal $0$, as the only term on the  is $1$

By the rule for multiplying power series, this gives:


 * $\dfrac {B_0} {0!} \dfrac 1 {n!} + \dfrac {B_1} {1!} \dfrac 1 {\left({n - 1}\right)!} + \dfrac {B_2} {2!} \dfrac 1 {\left({n - 2}\right)!} + \cdots + \dfrac {B_{n - 1} } {\left({n - 1}\right)!} \dfrac 1 {1!} = 0$

for that $n = 1$.

Multiplying through by $n!$:


 * $\dfrac {n!} {0! n!} B_0 + \dfrac {n!} {1! \left({n - 1}\right)!} B_1 + \dfrac {n!} {2!\left({n - 2}\right)!} B_2 + \cdots + \dfrac {n!} {\left({n - 1}\right)! 1!} B_{n - 1} = 0$

But those coefficients are the binomial coefficients:


 * $\dbinom n 0 B_0 + \dbinom n 1 B_1 + \dbinom n 2 B_2 + \cdots + \dbinom n {n - 1} B_{n - 1} = 0$

Hence the result.