Sum of Bernoulli Numbers by Binomial Coefficients Vanishes

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Theorem

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

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


Proof

Take the definition of Bernoulli numbers:

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


From the definition of the exponential function:

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

Thus:

\(\ds 1\) \(=\) \(\ds \paren {\frac x {e^x - 1} } \paren {\frac {e^x - 1} x}\)
\(\ds \) \(=\) \(\ds \paren {\sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!} } \paren {\sum_{n \mathop = 1}^\infty \frac {x^{n - 1} } {n!} }\)
\(\ds \) \(=\) \(\ds \paren {\sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!} } \paren {\sum_{n \mathop = 0}^\infty \frac {x^n} {\paren {n + 1}!} }\) as both series start at zero


By Product of Absolutely Convergent Series, we will let:

\(\ds a_n\) \(=\) \(\ds \frac {B_n x^n} {n!}\)
\(\ds b_n\) \(=\) \(\ds \frac {x^n} {\paren {n + 1}!}\)

Then:

\(\ds \sum_{n \mathop = 0}^\infty c_n\) \(=\) \(\ds \paren {\sum_{n \mathop = 0}^\infty a_n} \paren {\sum_{n \mathop = 0}^\infty b_n}\) \(\ds =1\)
\(\ds c_n\) \(=\) \(\ds \sum_{k \mathop = 0}^n a_k b_{n - k}\)
\(\ds c_0\) \(=\) \(\ds \frac {B_0 x^0} {0!} \frac {x^0} {\paren {0 + 1}!}\) \(\ds = 1\) as $c_0 = \paren {a_0} \paren {b_{0 - 0} } = \paren {a_0} \paren {b_0}$
\(\ds \leadsto \ \ \) \(\ds \sum_{n \mathop = 1}^\infty c_n\) \(=\) \(\ds \paren {\sum_{n \mathop = 0}^\infty a_n} \paren {\sum_{n \mathop = 0}^\infty b_n} - a_0 b_0\) \(\ds = 0\) subtracting $1$ from both sides
\(\ds \) \(=\) \(\ds c_1 x + c_2 x^2 + c_3 x^3 + \cdots\) \(\ds = 0\)
\(\ds \leadsto \ \ \) \(\ds \forall n \in \Z_{>0}: \, \) \(\ds c_n\) \(=\) \(\ds 0\)


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


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

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


But those coefficients are the binomial coefficients:

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

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}$


Also see


Sources

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