How is $ B_n = 1- \sum_{k=0}^{n-1} \binom{n}{k} \frac{B_k}{n-k+1} $ Where $B_n$ are the Bernoulli Numbers with $B_1 = \frac{1}{2}$

Question

So I was browing Wikipedia just looking at Bernoulli Number identities and I stumbled across this $$ B_m^+ = 1- \sum_{k=0}^{m-1} \binom{m}{k} \frac{B_k^+}{m-k+1} $$ The Wikipedia page said that this identity was derived from the identity : $$ \sum_{k=0}^{m}\left(\begin{array}{c} m+1 \\ k \end{array}\right) B_{k}^{+}=m+1 $$ My question is, how are these identities derived? Wikipedia gives the reference for these theorems to be from the page on Bernoulli Numbers on WolframMathWorld but I can't find the proof of any of these identities there. Can anyone guide me to or state a proof of either identity? (From the definition of Bernoulli Numbers stated below) $$ \frac{t}{1-e^{-t}}=\frac{t}{2}\left(\operatorname{coth} \frac{t}{2}+1\right) \quad=\sum_{m=0}^{\infty} \frac{B_{m}^{+} t^{m}}{m !} $$

Answer 1

As Wikipedia says, the first identity can be obtained from the second by "solving" the latter for $\ B_m^+\ $: \begin{align} m+1&=\sum_{k=0}^m{m+1\choose k}B_k^+\\ &=(m+1)B_m^++\sum_{k=0}^{m-1}\frac{(m+1)!}{k!\,(m+1-k)!}B_k^+\\ &=(m+1)\left(B_m^++\sum_{k=0}^{m-1}\frac{m!}{k!\,(m-k)!}\left(\frac{B_k^+}{m-k+1}\right)\right)\\ &=(m+1)\left(B_m^++\sum_{k=0}^{m-1}{m\choose k}\frac{B_k^+}{m-k+1}\right)\ . \end{align} Dividing through by $\ m+1\ $ and subtracting $\ \displaystyle\sum_{k=0}^{m-1}{m\choose k}\frac{B_k^+}{m-k+1}\ $ from both sides of this equation gives the required first identity.

Here is a version of reuns's derivation of the second identity which works directly from the identity $$ \frac{t}{1-e^{-t}}=\sum_{k=0}^\infty\frac{B_k^+t^k}{k!}\ . $$ Multiplying both sides of this identity by $\ e^t-1\ $ gives \begin{align} \sum_{m=0}^\infty\frac{t^{m+1}}{m!}&= te^t\\ &=\big(e^t-1\big)\sum_{k=0}^\infty\frac{B_k^+t^k}{k!}\\ &=\sum_{n=0}^\infty\sum_{k=0}^\infty\frac{B_k^+t^{n+k+1}}{(n+1)!k!}\\ &=\sum_{m=0}^\infty t^{m+1}\sum_{k=0}^m\frac{B_k^+}{(m+1-k)!k!}\ . \end{align} Equating coefficients of $\ t^{m+1}\ $ gives $$ \frac{1}{m!}=\sum_{k=0}^m\frac{B_k^+}{(m+1-k)!k!}\ , $$ and multiplying this equation by $\ (m+1)!\ $ gives the required identity.

Answer 2

The definition of $B_k$ is $$\frac{z}{e^z-1}=\sum_{k=0}^\infty \frac{B_k}{k!} z^k$$ it gives $$\sum_{m=0}^\infty \frac{z^{m+1}}{m!}=(e^z-1)\frac{-z}{e^{-z}-1}=\sum_{m=0}^\infty z^{m+1} \sum_{k=0}^m\frac{(-1)^k B_k}{k! (m-k+1)!} $$

Conclude by equating the coefficients and using $B_k^+ = (-1)^k B_k$.