# Bernoulli numbers

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**Bernoulli numbers** \(B_n\) are a sequence of rational numbers that satisfy the generating functions \[\displaystyle \dfrac t{e^t-1} = \sum_{m=0}^\infty B_m \dfrac{t^m}{m!}.\]
Bernoulli Numbers are also useful in finding the values of \(\zeta(n)\) for even \(n\)'s.
You may try this for its application.

The values of the first few Bernoulli numbers are as follows:

\[n\] | \[0\] | \[1\] | \[2\] | \[3\] | \[4\] | \[5\] | \[6\] | \[7\] | \[8\] | \[9\] | \[10\] |

\(B_n\) | \(1\) | \(\pm \frac 12 \) | \(\frac16 \) | \(0 \) | \(-\frac1{30} \) | \(0 \) | \(\frac1{42} \) | \(0 \) | \(-\frac1{30} \) | \(0 \) | \(\frac5{66} \) |

#### Contents

## Properties

\[B_{2k+1}=0 ~\forall k\ge 1.\ _\square\]

We see that \[\dfrac{t}{2}+\dfrac{t}{e^t-1}=\dfrac{t(e^t+1)}{2(e^t-1)}=\dfrac{t}{2}\coth\dfrac{t}{2}\] is an even function of \(t\).

So in its power series expansion about \(t = 0,\) the odd order coefficients are zero.

So \(B_1+\dfrac{1}{2},B_3,B_5,B_7,\ldots\) are zero.Hence, \(B_1=-\dfrac{1}{2}\) and \(B_{2k+1}=0 ~\forall k\ge 1.\ _\square\)

## See Also

**Cite as:**Bernoulli numbers.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/bernoulli-numbers/