# Sum of Bernoulli Numbers by Binomial Coefficients Vanishes

## Theorem

$\forall n \in \Z_{>1}: \displaystyle \sum_{k \mathop = 0}^{n - 1} \binom n k B_k = 0$

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:

 $\displaystyle e^x$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$ $\displaystyle$ $=$ $\displaystyle 1 + \sum_{n \mathop = 1}^\infty \frac {x^n} {n!}$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {e^x - 1} x$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty \frac {x^{n - 1} } {n!}$ $\displaystyle$ $=$ $\displaystyle 1 + \frac x {2!} + \frac {x^2} {3!} + \cdots$

Thus:

 $\displaystyle 1$ $=$ $\displaystyle \paren {\frac x {e^x - 1} } \paren {\frac {e^x - 1} x}$ $\displaystyle$ $=$ $\displaystyle \paren {\sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!} } \paren {\sum_{n \mathop = 1}^\infty \frac {x^{n - 1} } {n!} }$ $\displaystyle$ $=$ $\displaystyle \paren {\sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!} } \paren {\sum_{n \mathop = 0}^\infty \frac {x^{n} } {\paren{n + 1 }!} }$ Both series start at zero

By Product of Absolutely Convergent Series, we will let:

 $\displaystyle a_n$ $=$ $\displaystyle \frac {B_n x^n} {n!}$ $\displaystyle b_n$ $=$ $\displaystyle \frac {x^{n} } {\paren {n + 1 }!}$

Then:

 $\displaystyle \sum_{n \mathop = 0}^\infty c_n$ $=$ $\displaystyle \paren { \displaystyle \sum_{n \mathop = 0}^\infty a_n } \paren {\displaystyle \sum_{n \mathop = 0}^\infty b_n }$ $\displaystyle =1$ $\displaystyle c_n$ $=$ $\displaystyle \sum_{k \mathop = 0}^n a_k b_{n - k}$ $\displaystyle c_0$ $=$ $\displaystyle \frac {B_0 x^0} {0!} \frac {x^{0} } {\paren {0 + 1 }!}$ $\displaystyle = 1$ $c_0 = \paren {a_0 } \paren {b_{0 - 0 } } = \paren {a_0 } \paren {b_0 }$ $\displaystyle \leadsto \ \$ $\displaystyle \sum_{n \mathop = 1}^\infty c_n$ $=$ $\displaystyle \paren { \displaystyle \sum_{n \mathop = 0}^\infty a_n } \paren {\displaystyle \sum_{n \mathop = 0}^\infty b_n } - a_0 b_0$ $\displaystyle =0$ Subtract 1 from both sides of the equation. $\displaystyle$ $=$ $\displaystyle c_1x + c_2x^2 + c_3x^3 + \cdots$ $\displaystyle =0$

$\leadsto \forall n \in \Z_{\gt 0}$, $c_n = 0$.

 $\displaystyle c_1$ $=$ $\displaystyle \frac {B_0 x^0} {0!} \frac {x^{1} } {\paren {1 + 1 }!} + \frac {B_1 x^1} {1!} \frac {x^{0} } {\paren {0 + 1 }!}$ $\displaystyle = 0$ $= a_0 b_1 + a_1 b_0$ $\displaystyle c_2$ $=$ $\displaystyle \frac {B_0 x^0} {0!} \frac {x^{2} } {\paren {2 + 1 }!} + \frac {B_1 x^1} {1!} \frac {x^{1} } {\paren {1 + 1 }!} + \frac {B_2 x^2} {2!} \frac {x^{0} } {\paren {0 + 1 }!}$ $\displaystyle = 0$ $= a_0 b_2 + a_1 b_1 + a_2 b_0$ $\displaystyle \cdots$ $=$ $\displaystyle \cdots$ $\displaystyle = 0$ $\displaystyle c_n$ $=$ $\displaystyle \frac {B_0 x^0} {0!} \frac {x^{n} } {\paren {n + 1 }!} + \frac {B_1 x^1} {1!} \frac {x^{n-1} } {\paren {n - 1 + 1 }!} + \cdots + \frac {B_n x^n} {n!} \frac {x^{0} } {\paren {0 + 1 }!}$ $\displaystyle = 0$ $= a_0 b_n + a_1 b_{n - 1 } + a_2 b_{n - 2 } + \cdots + a_n b_0$

Multiplying $c_n$ through by $\paren {n + 1 }!$ gives:

 $\displaystyle \paren {n + 1 }! c_n$ $=$ $\displaystyle \frac {B_0 x^0} {0!} \frac {\paren {n + 1 }! x^n } {\paren {n + 1 }!} + \frac {B_1 x^1} {1!} \frac {\paren {n + 1 }! x^{n-1} } {\paren {n - 1 + 1 }!} + \cdots + \frac {B_n x^n} {n!} \frac {\paren {n + 1 }! x^{0} } {\paren {0 + 1 }!}$ $\displaystyle = 0$ $\displaystyle$ $=$ $\displaystyle x^n \paren {\frac {\paren {n + 1 }! } {0! \paren {n + 1 }!} B_0 + \frac {\paren {n + 1 }! } {1! \paren {n - 1 + 1 }!} B_1 + \cdots + \frac {\paren {n + 1 }! } {n! \paren {0 + 1 }!} B_n }$ $\displaystyle = 0$ factoring out $x^n$

But those coefficients are the binomial coefficients:

 $\displaystyle \paren {n + 1 }! c_n$ $=$ $\displaystyle \dbinom {n + 1 } 0 B_0 + \dbinom {n + 1 } 1 B_1 + \dbinom {n + 1 } 2 B_2 + \cdots + \dbinom {n + 1 } n B_n$ $\displaystyle = 0$ $\displaystyle n! c_{n-1 }$ $=$ $\displaystyle \dbinom n 0 B_0 + \dbinom n 1 B_1 + \dbinom n 2 B_2 + \cdots + \dbinom n {n - 1} B_{n - 1}$ $\displaystyle = 0$

Hence the result.

$\blacksquare$

## Examples

$\begin{array}{r|cccccccccc} B_k & \dbinom n 0 & & \dbinom n 1 & & \dbinom n 2 & & \dbinom n 3 & & \dbinom n 4 & & \dbinom n 5 \\ \hline B_0 = 1 & 1 B_0 & & & & & & & & & & & = 1 \\ B_1 = -\frac 1 2 & 1 B_0 & + & 2 B_1 & & & & & & & & & = 0 \\ B_2 = +\frac 1 6 & 1 B_0 & + & 3 B_1 & + & 3 B_2 & & & & & & & = 0 \\ B_3 = 0 & 1 B_0 & + & 4 B_1 & + & 6 B_2 & + & 4 B_3 & & & & & = 0 \\ B_4 = -\frac 1 {30} & 1 B_0 & + & 5 B_1 & + & 10 B_2 & + & 10 B_3 & + & 5 B_4 & & & = 0 \\ B_5 = 0 & 1 B_0 & + & 6 B_1 & + & 15 B_2 & + & 20 B_3 & + & 15 B_4 & + & 6 B_5 & = 0 \\ \end{array}$