Equivalence of Definitions of Bernoulli Numbers

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Theorem

The following definitions of the concept of Bernoulli Numbers are equivalent:

Generating Function

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

Recurrence Relation

$B_n = \begin{cases} 1 & : n = 0 \\ \ds - \sum_{k \mathop = 0}^{n - 1} \binom n k \frac {B_k} {n + 1 - k} & : n > 0 \end{cases}$


or equivalently:

$B_n = \begin{cases} 1 & : n = 0 \\ \ds - \frac 1 {n + 1} \sum_{k \mathop = 0}^{n - 1} \binom {n + 1} k B_k & : n > 0 \end{cases}$


Proof



From the generating function definition:

\(\ds \frac x {e^x - 1}\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!}\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!} \paren {\sum_{k \mathop = 0}^\infty \frac {x^k} {k!} - 1}\) Definition of Real Exponential Function
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!} \sum_{k \mathop = 0}^\infty \frac {x^{k + 1} } {\paren {k + 1}!}\) $1 = \dfrac {x^0} {0!}$
\(\ds \leadsto \ \ \) \(\ds 1\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!} \sum_{k \mathop = 0}^\infty \frac {x^k} {\paren {k + 1}!}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!} \sum_{k \mathop = 0}^n \binom n k \frac {B_k} {n - k + 1}\) Cauchy Product, Product of Absolutely Convergent Series


Equating coefficients:

For $n = 0$:

\(\ds 1\) \(=\) \(\ds \sum_{k \mathop = 0}^0 \binom 0 k \frac {B_k} {0 - k + 1}\)
\(\ds \) \(=\) \(\ds \binom 0 0 \frac {B_0} {0 - 0 + 1}\)
\(\ds \) \(=\) \(\ds B_0\) Binomial Coefficient with Zero


For $n > 0$:

\(\ds 0\) \(=\) \(\ds \frac 1 {n!} \sum_{k \mathop = 0}^n \binom n k \frac {B_k} {n - k + 1}\)
\(\ds \) \(=\) \(\ds \frac 1 {n!} \paren {\sum_{k \mathop = 0}^{n - 1} \binom n k \frac {B_k} {n - k + 1} + \binom n n \frac {B_n} {n - n + 1} }\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^{n - 1} \binom n k \frac {B_k} {n - k + 1} + B_n\) Binomial Coefficient with Self and simplifying
\(\ds \leadsto \ \ \) \(\ds B_n\) \(=\) \(\ds -\sum_{k \mathop = 0}^{n - 1} \binom n k \frac {B_k} {n - k + 1}\)


Hence the result:

$B_n = \begin{cases} 1 & : n = 0 \\ \ds - \sum_{k \mathop = 0}^{n-1} \binom n k \frac {B_k} {n - k + 1} & : n > 0 \end{cases}$

$\blacksquare$