A Bernoulli Number Identity

The point of this note is to record a little observation about bernoulli numbers via thinking about the $\zeta$ function and guess-and-checking some empirically occurring patterns.

It looks incredibly similar to Faulhaber's formula, except I think it would require me substituting $1/2$ for $n$, which is supposed to be an integer. Anyway this similarity makes me suspect it should have a pretty elementary (and probably well-known) proof, but I don't see it quite yet.

Multiplying the $\zeta$ function

If I have as usual \[ \zeta(s) = \sum_{0 < i} {1\over i^s} = 1 + {1\over 2^s} + {1\over 3^s} + {1\over 4^s} + \cdots\] Then observe \[ \zeta(s)\zeta(t) = \left( \green {1\over 1^{s+t}} + \red{{1\over2^s 1^t } + {1\over 3^s1^t} + {1\over 4^s1^t } + \cdots} \right.\] \[ \qquad\qquad\qquad\qquad + \blue{1\over 1^s 2^t} + \green{1\over 2^{s+t}} + \red{{1\over3^s 2^t } + {1\over4^s 2^t} + \cdots} \] \[ \qquad\qquad\qquad\qquad + \blue{{1\over 1^s 3^t} + {1\over 2^s 3^t }} + \green{1\over 3^{s+t}} + \red{{1\over 4^s3^t} + \cdots} \] \[ \qquad\qquad\qquad\qquad \left. + \blue{{1\over 1^s 4^t} + {1\over 2^s 4^t } + {1\over 3^s 4^t}} + \green{1\over 4^{s+t}} + \red \cdots \right)\] could be written \[ \zeta(s)\zeta(t) = \blue{ \sum_{0 < i < j} {1\over i^s j^t}} + \green{\sum_{0 < i} {1\over i^{s+t}}} + \red{ \sum_{0 < j < i} {1\over i^s j^t}} \tag {*}\] This suggests a generalization of the $\zeta$ function to \[\zeta(s, t) = \sum_{0 < i < j} {1\over i^s j^t}\] in which case we can massage $(*)$ into \[ \zeta(s)\zeta(t) = \zeta(t,s) + \zeta(s+t) + \zeta(s,t) = 2\zeta(s,t) + \zeta(s+t) \] Knowing that $\zeta(2) = \pi^2/6$ and $\zeta(4) = \pi^4 = 90$, we might try plugging in $2$ for $s$ and $t$, to get \[ \zeta(2)\zeta(2) = 2\zeta(2,2) + \zeta(4) \] hence \[ \zeta(2,2) = {\zeta(2)^2 - \zeta(4) \over 2} = {\pi^4\over 2}\left({1\over 36} - {1\over 90}\right) = {\pi^4\over 120}\] That's surprisingly nice! 120 is a nice round value that happens to be 5 factorial.

Generalizing the multi-$\zeta$

Suppose we define \[\zeta(s_1, s_2, \ldots, s_n) = \sum_{0 < i_1 < \cdots < i_n} {1\over i_1^{s_1}\cdots i_n^{s_n}}\] and set ourselves the goal of trying to compute $\zeta(2,2,\ldots 2)$ in general.

There is a family of identities that arise from multiplying a "multi-$\zeta$" function by a single one. \[\zeta(a) \zeta(\rx) = \zeta(\rx, a) + \zeta(a+\rx) + \zeta(a, \rx)\] \[\zeta(a, b) \zeta(\rx) = \zeta(\rx, a, b) + \zeta(a+\rx, b) + \zeta(a, \rx, b) + \zeta(a, b+\rx) + \zeta(a,b,\rx) \] \[\zeta(a, b, c) \zeta(\rx) = \zeta(\rx, a, b, c) + \zeta(a+\rx, b, c) + \zeta(a, \rx, b, c) + \zeta(a, b+\rx, c) + \] \[\zeta(a, b, \rx, c) + \zeta(a, b, c+\rx) + \zeta(a, b, c, \rx)\] etc.

These are established by reasoning similar to that above. In every case we need only account for how the $x$ is "inserted" into the existing ordered list of variables, possibly colliding with existing variables.

Now define \[\zeta_k(a) = \zeta(\overbrace{a, \cdots, a}^{k}) \qquad \zeta_0(a) = 1\] \[\zeta_k(a \sim b) = \zeta(b, \overbrace{a, \cdots, a}^{k-1})+\zeta(a, b, \overbrace{a, \cdots, a}^{k-2}) +\cdots +\zeta( \overbrace{a, \cdots, a}^{k-1}, b) \] What we want to compute is $\zeta_k(2)$. Observe that \[\zeta_k(a\sim a) = k \zeta_k(a)\] and \[\zeta_k(a)\zeta(ma) = \zeta_{k+1}(a \sim ma) + \zeta_k(a\sim (m+1)a)\] which means \[\zeta_{k}(a \sim ma) = \zeta_{k-1}(a)\zeta(ma) - \zeta_{k-1}(a\sim (m+1)a) \] and if we set $m=1$ this telescopes out to \[\zeta_{k}(a \sim a) = \zeta_{k-1}(a)\zeta(a) - \zeta_{k-2}(a)\zeta(2a) + \zeta_{k-3}(a)\zeta(3a) - \cdots \] that is, \[\zeta_{k}(a) = {\zeta_{k-1}(a)\zeta(a) - \zeta_{k-2}(a)\zeta(2a) + \zeta_{k-3}(a)\zeta(3a) - \cdots \pm \zeta(ka) \over k} \tag{\dag}\] So for example we find \[\zeta_{2}(a) = {\zeta(a)\zeta(a) - \zeta(2a)\over 2} \] \[\zeta_{3}(a) = {\zeta_2(a)\zeta(a) - \zeta(a)\zeta(2a) + \zeta(3a)\over 3} \] \[\zeta_{4}(a) = {\zeta_3(a)\zeta(a) - \zeta_2(a)\zeta(2a) + \zeta(a)\zeta(3a) - \zeta(4a)\over 4} \]

Concrete Values of $\zeta_k(2)$

Armed with some known values of $\zeta(2n)$, we compute and notice an interesting pattern: \[\zeta_{1}(2) = \zeta(2) = {\pi^2\over 3!}\] \[\zeta_{2}(2) = {\zeta(2)\zeta(2) - \zeta(4)\over 2} = {1\over 2}\left(\left(\pi^2\over 6\right)^2 - {\pi^4\over 90}\right) = {\pi^4\over 5!}\] \[\zeta_{3}(2) = {\zeta_2(2)\zeta(2) - \zeta(2)\zeta(4) + \zeta(6)\over 3} = {1\over 3}\left({\pi^4\over 5!}{\pi^2\over 6} -{\pi^2\over 6} {\pi^4\over 90} + {\pi^6\over 945}\right) = {\pi^6\over 7!}\] \[\zeta_{4}(2) = {\zeta_3(2)\zeta(2) - \zeta_2(2)\zeta(4) + \zeta(2)\zeta(6) - \zeta(8)\over 4} =\] \[ {1\over 4}\left({\pi^6\over 7!}{\pi^2\over 6} -{\pi^4\over 5!} {\pi^4\over 90} + {\pi^2\over 6}{\pi^6\over 945} - {\pi^8\over 9450}\right) = {\pi^8\over 9!} \] Therefore we conjecture that, for all $k \in \N$ that we have \[\zeta_{k}(2) = {\pi^{2k}\over (2k+1)!}\]

Manipulations

Notice that $(\dag)$ is equivalent to \[M \zeta_M(a) = \sum_{n=1}^M (-1)^{n+1} \zeta_{M-n}(a) \zeta(an)\] It would therefore suffice to prove the conjecture if we could prove it inductively along this recurrence for the particular value $a = 2$, i.e. if we could show \[M {\pi^{2M}\over (2M+1)!} = \sum_{n=1}^M (-1)^{n+1} {\pi^{2(M-n)}\over (2(M-n)+1)!} \zeta(2n)\] But let's plug in the known expression for the $\zeta$ function at even integers, namely \[ \zeta(2n) = (-1)^{n+1} {2^{2n-1} \pi^{2n} B_{2n}\over (2n)!}\] and we get \[M {\pi^{2M}\over (2M+1)!} = \sum_{n=1}^M (-1)^{n+1} {\pi^{2(M-n)}\over (2(M-n)+1)!} (-1)^{n+1} {2^{2n-1} \pi^{2n} B_{2n}\over (2n)!}\] which simplifies to \[M {\pi^{2M}\over (2M+1)!} = \sum_{n=1}^M {\pi^{2M}\over (2(M-n)+1)!} {2^{2n-1} B_{2n}\over (2n)!}\] and we can cancel the $\pi$s and move the factorial over to the right: \[M = \sum_{n=1}^M {2M+1 \choose 2n} {2^{2n-1} B_{2n}}\] Since odd Bernoulli numbers for $n \ge 3$ are zero, this is the same thing as \[M = \sum_{n=2}^{2M} {2M+1 \choose n} {2^{n-1} B_{n}}\] and by adding some terms to adjust the sum bounds we see \[\left({2M+1 \choose 0} {2^{-1} B_{0}}\right) + \left({2M+1 \choose 1} {2^{0} B_{1}}\right) + M = \sum_{n=0}^{2M} {2M+1 \choose n} {2^{n-1} B_{n}}\] and so \[\left(1\cdot (1/ 2) \cdot 1\right) + \left((2M+1)\cdot 1 \cdot (-1/2)\right) + M = \sum_{n=0}^{2M} {2M+1 \choose n} {2^{n-1} B_{n}}\] and then the entire left side of the equation cancels, leaving \[0 = \sum_{n=0}^{2M} {2M+1 \choose n} {2^{n-1} B_{n}}\] which we can tidy up if we want by multiplying by 2: \[\sum_{n=0}^{2M} {2M+1 \choose n} {2^{n} B_{n}} = 0\] Now $B_{2M+1}$ is zero if $M \ge 1$, so in fact \[\sum_{n=0}^{2M+1} {2M+1 \choose n} {2^{n} B_{n}} = 0\] which means we can equvialently say

Final Form

Conjecture For any odd positive $M$, we have \[\sum_{n=0}^{M} {M \choose n} {2^{n} B_{n}} = 0\] This can be empirically verified, e.g. on sagemath with code like

[sum(bernoulli(n) * 2^n * binomial(2*M+1, n) for n in range(0, 2*M+2)) for M in range(1, 10)]

which evaluates to a vector of all zeroes.

Postscript

I notice that apparently the expression \[2\left(\left(\sum_{n=0}^{M} {M \choose n} {2^{n} B_{n}}\right) - B_M\right)\] generates the Genocchi numbers of the first kind, that is,

[2 * (sum(bernoulli(n) * 2^n * binomial(M, n) for n in range(0, M+1)) - bernoulli(M)) for M in range(1, 15)]

yields

[1, -1, 0, 1, 0, -3, 0, 17, 0, -155, 0, 2073, 0, -38227]

Hm, what I have is very close to the formula

a(n) = Sum_{k=0..n-1} binomial(n,k) 2^k*B(k)

listed in that oeis link. I think this means \[\sum_{n=0}^{M} {M \choose n} {2^{n} B_{n}} = (2-2^M)B_M\] as confirmed by

print([sum(bernoulli(n) * 2^n * binomial(M,n) for n in range(0,M+1)) - bernoulli(M) * (2-2^M) for M in range(0,10)])

printing all zeroes.