Denote by $B_n$ the Bernoulli sequence (defined by the exponential generating function $\frac{x}{e^x-1}=\sum_{n=0}^{\infty}\frac{B_n}{n!}x^n$). As we know $$ \sum^{n}_{j=0}\binom{n}{j}B_j=B_n\; ; \; n\geq 2 $$ what about $\sum^{n}_{j=0}(-1)^j\binom{n}{j}B_j$, and as a more general case $$ \sum^{n}_{j=0}\frac{(-1)^{j+1-k}}{j+1}\binom{n}{j}\binom{j+1}{k}B_{j+1-k} $$ where $k$ is a given integer such that $1\leq k\leq n+1$ (is there any similar identity?). Note that putting $k=1$ […]

As we know the closed form of $$ {1 \over k} {n \choose k – 1} \sum_{j = 0}^{n + 1 – k}\left(-1\right)^{\, j} {n + 1 – k \choose j}B_{j} = \bbox[10px,border:2px solid #00A000]{{n! \over k!}\, \left[\, z^{n + 1 – k}\,\,\,\right] {z \exp\left(z\right) \over 1 – \exp\left(-z\right)}} $$ where $B_{n}$ the Bernoulli sequence […]

Is there anyone knows where is some official reference mentioning the relationship between cumulants of uniform distribution and the Bernoulli numbers (http://en.wikipedia.org/wiki/Cumulant#Cumulants_of_some_continuous_probability_distributions)?

The Bernoulli numbers were being used long before Bernoulli wrote about them, but according to Wikipedia, “The Swiss mathematician Jakob Bernoulli (1654–1705) was the first to realize the existence of a single sequence of constants B0, B1, B2, … which provide a uniform formula for all sums of powers.” Did he publish an exponential generating […]

The exercise reads “Express the power series for $\large \frac{z}{\sin (z)} = \frac{2 i z}{e^{iz} – e^{-iz}} $ in terms of Bernoulli numbers.” I am given in a previous exercise that the Bernoulli numbers are defined by $$ \frac{z}{e^z – 1} = \sum_{n = 0}^{\infty} \frac{B_n}{n!} z^n, $$ where $ B_n $ is the $ […]

The recursion formula $$\sum_{k=0}^{n-1}{n\choose k}B_k=0$$ which is equation (34) from the MathWorld page is the most basic bernoulli number recursion. Another recursion is given by the following:Consider $$\mathbf{A} = \begin{bmatrix} -\frac{1}{2} & -\frac{1}{6} & -\frac{1}{12} & -\frac{1}{20} & -\frac{1}{30} & -\frac{1}{42}\\ \frac{1}{1} & 0 & 0 & 0 & 0 & 0\\ 0 & \frac{1}{2} […]

I’ve read that Bernoulli Numbers are defined by the series $$ \frac{z}{e^z-1}\equiv \sum\limits_{n=0}^{\infty}B_n\frac{z^n}{n!},$$ So if I start with $0$ I get $$ B_0\frac{1}{1}=B_0{1}. $$ My question is, why is there a $B_0$ in the term…is it of any significance? Or just a “marker” or something to indicate that this is the $B_0$ term? If I […]

Question : Is $$\frac{\zeta (m+n)}{\zeta (m)\zeta (n)}$$ a rational number for $m,n\ge 2\in\mathbb N$ where $\zeta (s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$? Motivation : We know that $$\zeta (2k)=(-1)^{k+1}\frac{B_{2k}(2\pi)^{2k}}{2(2k)!}$$ and that $B_{2k}$ is a rational number for any $k\in\mathbb N$ where $B_n$ is the Bernoulli numbers. Hence, if both $m$ and $n$ are even, then we can see that $$\frac{\zeta […]

A recent SciComp.SE Question motivates us to ask for a nice continued fraction expansion of the following Maclaurin series: $$ f(x) = \sum_{n=0}^\infty \frac{B_n\; x^{n+3}}{n! (n+3)} = \int_0^x \frac{t^3}{e^t – 1} dt $$ where $B_n$ are the (first) Bernoulli numbers. Although the positive power of $t$ in the numerator of the integrand removes the singularity […]

Consider a sequence of $n$ Bernoulli trials with $P(\text{success})=p$. Let $X_i$ and $X_j$ be indicator variables of the number of “success” in $i$th and $j$th runs. Given the total number of success was $m$, $m<n$. I am asked to compute the correlation coefficient for $X_i$ and $X_j$. Now I know the formula for correlation coefficient, […]

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