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[parent] the odd Bernoulli numbers are zero (Theorem)

Recall that, for $ k\geq 0$, the Bernoulli numbers $ B_k$ are defined as the coefficients in the Taylor expansion:

$\displaystyle \frac{t}{e^t-1}=\sum_{k\geq 0} B_k \frac{t^k}{k!}.$     (1)

Just to name a few:
$\displaystyle B_0=1,\quad B_1=-\frac{1}{2},\quad B_2=\frac{1}{6},\quad B_3=0,\quad B_4=-\frac{1}{30},\ B_5=0,\ldots,\ B_{10}=\frac{5}{66},\ldots $
Lemma 1   If $ k\geq 3$ is odd then $ B_k=0$.
Proof. From the right hand side of (1) we extract the term corresponding to $ k=1$:
$\displaystyle \frac{t}{e^t-1}=-\frac{t}{2}+\sum_{k\geq 0,\ k\neq 1} B_k \frac{t^k}{k!}.$     (2)

Thus:
$\displaystyle \frac{t}{e^t-1}+\frac{t}{2}=\sum_{k\geq 0,\ k\neq 1} B_k \frac{t^k}{k!}$     (3)

and the left hand side can be rewritten as:
$\displaystyle \frac{t}{e^t-1}+\frac{t}{2}= \frac{2t+t(e^t-1)}{2(e^t-1)} = \frac... ...frac{e^t+1}{e^t-1}=\frac{t}{2} \cdot \frac{e^{t/2}+e^{-t/2}}{e^{t/2}-e^{-t/2}}.$     (4)

Hence, if one replaces $ t$ by $ -t$ then (4) is unchanged. Since (4) is the left hand side of (3), the quantity
$\displaystyle \sum_{k\geq 0,\ k\neq 1} B_k \frac{t^k}{k!}$
is also unchanged when $ t$ is exchanged by $ -t$, and so we must have $ B_k=(-1)^kB_k$ for $ k\neq 1$. We conclude that if $ k\geq 3$ and $ k$ is odd, $ B_k=0$. $ \qedsymbol$



"the odd Bernoulli numbers are zero" is owned by alozano.
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See Also: Kummer's congruence, congruence of Clausen and von Staudt

Keywords:  Bernoulli number

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Cross-references: left hand side, term, right hand side, odd, Taylor expansion, coefficients, Bernoulli numbers
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This is version 2 of the odd Bernoulli numbers are zero, born on 2005-04-20, modified 2005-04-20.
Object id is 6959, canonical name is OddBernoulliNumbersAreZero.
Accessed 1425 times total.

Classification:
AMS MSC: 11B68 (Number theory :: Sequences and sets :: Bernoulli and Euler numbers and polynomials)
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