Stackexchange isn't getting really excited about this, so here it is.

The $n$th cumulant of the uniform probability distribution on the interval $[-1,0]$ is $B_n/n$, where $B_n$ is the $n$th Bernoulli number.

And $-\zeta(1-n)=B_n/n$, where $\zeta$ is Riemann's function.

Those two facts can be derived, but is there some argument that shows, without doing that, that you'd *expect* the cumulants of the uniform distribution to be those values of the $\zeta$ function?

Might this shed some light either on the $\zeta$ function or the uniform distribution? Or anything else?

**Appendix:** Cumulants are like moments, but better. The $n$th cumulant of a probability distribution is homogeneous of degree $n$ (like the $n$th moment). If $n=1$ it is shift-equivariant; if $n>1$ it is shift-invariant (like the $n$th *central* moment). If $X_1,X_2,X_3,\ldots$ are independent random variables, then the $n$th cumulant of the distribution of their sum is just the sum of the $n$th cumulants of their distributions. That last property is shared by central moments only in the cases $n=2,3$ (where the cumulant is just the central moment). Simplest nontrivial example: the 4th cumulant is the fourth central moment minus 3 times the square of the second central moment.

**Later addition (should this be a separate question?):** The number of independent Bernoulli trials strictly preceding the first success, with probability $1/(1+c)$ of success on each trial, is a geometrically distributed random variable with expected value $c$, taking values in the set $\lbrace 0,1,2,\ldots \rbrace$. The first several cumulants of that distribution are these
$$
\begin{align}
& c \\
& c+c^2 \\
& c + 3c^2 + 2c^3 \\
& c + 7c^2 + 12c^3 + 6c^4 \\
& c + 15c^2 + 50c^3 + 60c^4 + 24c^5 \\
& c + 32c^2 +180c^3+390c^4+360c^5+120c^6
\end{align}
$$
The probability distribution of the number of successes in just one trial, with probability $c$ of success on each trial, has cumulants
$$
\begin{align}
& c \\
& c-c^2 \\
& c - 3c^2 + 2c^3 \\
& c - 7c^2 + 12c^3 - 6c^4 \\
& c - 15c^2 + 50c^3 - 60c^4 + 24c^5 \\
& c - 32c^2 +180c^3-390c^4+360c^5-120c^6
\end{align}
$$
They're the same except for alternating signs.

If it were possible for the probability to be $-c$, where $c>0$, this sequence would be just $-1$ times the sequence of cumulants of the geometric distribution. Likewise, the sequence of values of $\zeta(1-n)$ is $-1$ times the sequence of cumulants of the probability distribution specified above. Could $\zeta(1-n),\quad n=1,2,3,\ldots$ be the sequence of cumulants of a distribution that would exist if we allowed negative probabilities (but still required the measure of the whole probability space to be $1$)?

If so, do we have two instances of a phenomenon that can be stated in a general way?