Engineering Mathematics: Bernoulli-Zeta functions

BZ.ID.1: Bernoulli-Zeta relation

$\zeta (1-k)=-\frac{B_{k}}{k}$

Proof

$\zeta (1-k)=\sum _{n=1}^{\infty }\frac{1}{n^{1-k}}=\sum _{n=1}^{\infty }n^{k-1}$

(1)

based on the following identity

$\left ( \frac{d}{dt} \right )^{k-1}e^{nt} \mid _{t=0}=n^{k-1}$

(2)

substituting (2) into (1)

$\zeta (1-k)=\sum _{n=1}^{\infty }\left ( \frac{d}{dt} \right )^{k-1}e^{nt} \mid _{t=0}$

(3)

$\zeta (1-k)=\left ( \frac{d}{dt} \right )^{k-1}\sum _{n=1}^{\infty }e^{nt} \mid _{t=0}$

(4)

$\zeta (1-k)=\left ( \frac{d}{dt} \right )^{k-1}\left (\frac{1}{1-e^{t}} -1 \right ) \mid _{t=0}$

derivative of constant , which is 1, equals = 0

(5)

$\zeta (1-k)=\left ( \frac{d}{dt} \right )^{k-1}\left (\frac{1}{1-e^{t}} \right ) \mid _{t=0}$

(6)

$\zeta (1-k)=\left ( \frac{d}{dt} \right )^{k-1}\left (\frac{-1}{t}\sum _{n=1}^{\infty }B_{n}\frac{t^{n}}{n!} \right ) \mid _{t=0}$

(7)

$\zeta (1-k)=\left ( \frac{d}{dt} \right )^{k-1}\sum _{n=1}^{\infty }\left ( \frac{-B_{n}}{n} \right )\frac{t^{n-1}}{(n-1)!} \mid _{t=0}$

(8)

$\zeta (1-k)=-\frac{B_{k}}{k}$

if=k=2k

$\zeta (1-2k)=-\frac{B_{2k}}{2k}$

if n=k+1

$\zeta (-n)=\frac{B_{n+1}}{n+1}$

Engineering Mathematics