I would like to prove that $$B_{2n}=\frac{1}{2(2n-1)(2^{2n}-1)}\left(1-\sum_{k=1}^{n-1}(2k-1)2^{2k}\binom{2n}{2k}B_{2k}\right)$$ I've checked that it holds for n=2-20. This sum came up in a pet project I have with rational approximations, which gave me good reason to believe it; checking that it held by computing the sums was just a reassurance. Does anyone have suggestions on how to go about proving this?
Though the Bernoulli numbers and binomial coefficients seem to show up together in sum formulae fairly often, I didn't find any quite like this.
Working with the sum we find
$$\sum_{k=1}^{n-1} (2k-1) 2^{2k} {2n\choose 2k} B_{2k} \\ = 1 - (2n-1) 2^{2n} B_{2n} + \sum_{k=0}^{n} (2k-1) 2^{2k} {2n\choose 2k} B_{2k} \\ = 1 - (2n-1) 2^{2n} B_{2n} + \sum_{k=0}^{2n} (k-1) 2^{k} {2n\choose k} B_{k}.$$
Here we have used the fact that the only non-zero odd index Bernoulli number is $B_1$ and it gets canceled by the factor $k-1.$ Continuing with the sum term for the moment we find
$$\sum_{k=0}^{2n} (2n-k-1) 2^{2n-k} {2n\choose 2n-k} B_{2n-k} \\ = \sum_{k=0}^{2n} (2n-k-1) 2^{2n-k} {2n\choose 2n-k} (2n-k)! [z^{2n-k}] \frac{z}{\exp(z)-1} \\ = (2n)! [z^{2n}] \frac{z}{\exp(z)-1} \sum_{k=0}^{2n} (2n-k-1) 2^{2n-k} \frac{z^k}{k!}.$$
Evidently we get no contribution to the coefficient extractor from the sum when $k\gt 2n$ so we may write
$$ 2^{2n} (2n)! [z^{2n}] \frac{z}{\exp(z)-1} \sum_{k\ge 0} (2n-k-1) 2^{-k} \frac{z^k}{k!} \\ = 2^{2n} (2n)! [z^{2n}] \frac{z}{\exp(z)-1} \left((2n-1)\exp(z/2) - \sum_{k\ge 1} k 2^{-k} \frac{z^k}{k!}\right) \\ = 2^{2n} (2n)! [z^{2n}] \frac{z}{\exp(z)-1} \left((2n-1)\exp(z/2) - \frac{z}{2} \sum_{k\ge 1} 2^{-(k-1)} \frac{z^{k-1}}{(k-1)!}\right) \\ = 2^{2n} (2n)! [z^{2n}] \frac{z}{\exp(z)-1} \left((2n-1)\exp(z/2) - \frac{z}{2} \exp(z/2)\right) \\ = 2^{2n} (2n)! [z^{2n-1}] \frac{1}{\exp(z)-1} \left((2n-1)\exp(z/2) - \frac{z}{2} \exp(z/2)\right).$$
Observe that
$$[z^{2n-1}] (2n-1) \frac{1}{\exp(z)-1} \exp(z/2) = [z^{2n-1}] z \left( \frac{1}{\exp(z)-1} \exp(z/2)\right)' = [z^{2n-1}] z \left( \frac{1}{2} \frac{1}{\exp(z)-1} \exp(z/2) - \frac{\exp(z)}{(\exp(z)-1)^2} \exp(z/2)\right) = [z^{2n-1}] \frac{1}{\exp(z)-1} \left( \frac{z}{2} \exp(z/2) - \frac{z \exp(z)}{\exp(z)-1} \exp(z/2)\right).$$
It follows that the closed form for our sum is
$$- 2^{2n} (2n)! [z^{2n-1}] \frac{z \exp(z)}{(\exp(z)-1)^2} \exp(z/2) \\ = - 2^{2n} (2n)! [z^{2n}] \frac{z^2 \exp(3z/2)}{(\exp(z)-1)^2}.$$
This is analytic in a neigborhood of the origin with the pole there being canceled and hence the Cauchy Coefficient Formula applies. We find
$$- \frac{2^{2n} (2n)!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{2n+1}} \frac{z^2 \exp(3z/2)}{(\exp(z)-1)^2} \; dz \\ = - \frac{2^{2n} (2n)!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{2n-1}} \frac{\exp(3z/2)}{(\exp(z)-1)^2} \; dz.$$
Residues sum to zero and the residue at infinity is zero as well. We have the expansion
$$\frac{1}{(\exp(z)-1)^2} = \frac{1}{(z\pm 2\pi i k)^2} - \frac{1}{z\pm 2\pi i k} + \cdots$$
Therefore the residues at $\pm 2\pi i k$ with $k\ge 1$ require the derivative
$$\left(\frac{1}{z^{2n-1}} \exp(3z/2)\right)' = - \frac{2n-1}{z^{2n}} \exp(3z/2) + \frac{3}{2} \frac{1}{z^{2n-1}} \exp(3z/2).$$
Evaluate to obtain
$$ - \frac{2n-1}{(\pm 2\pi i k)^{2n}} \exp(\pm 3\pi i k) + \frac{3}{2} \frac{1}{(\pm 2\pi i k)^{2n-1}} \exp(\pm 3\pi i k)$$
The second term cancels when we add the pair of residues and we get
$$-2 \times (2n-1) \frac{(-1)^k}{2^{2n} \pi^{2n} (-1)^n k^{2n}}.$$
We also get from the second term of the expansion of $1/(\exp(z)-1)^2$
$$-\frac{1}{(\pm 2\pi i k)^{2n-1}} \exp(\pm 3\pi i k)$$
which cancels as well when paired.
Carrying out the necessary sign flip we thus obtain:
$$2 (2n-1) \frac{(-1)^n}{2^{2n}\pi^{2n}} \sum_{k\ge 1} \frac{(-1)^k}{k^{2n}} = - 2 (2n-1) \frac{(-1)^n}{2^{2n}\pi^{2n}} \left(1 - \frac{2}{2^{2n}}\right) \sum_{k\ge 1} \frac{1}{k^{2n}} \\ = - 2 (2n-1) \frac{(-1)^n}{2^{2n}\pi^{2n}} \left(1 - \frac{2}{2^{2n}}\right) \zeta(2n).$$
We return to the coefficient integral and evaluate the zeta function term by its standard closed form to get
$$2^{2n} (2n)! \times 2 (2n-1) \frac{(-1)^n}{2^{2n}\pi^{2n}} \left(1 - \frac{2}{2^{2n}}\right) \frac{(-1)^{n+1} B_{2n} (2\pi)^{2n}}{2(2n)!} \\ = - 2^{2n} (2n-1) \frac{1}{2^{2n}\pi^{2n}} \left(1 - \frac{2}{2^{2n}}\right) B_{2n} (2\pi)^{2n} \\ = (2n-1) (2-2^{2n}) B_{2n}.$$
We finally obtain for the initial sum term from the very start the closed form
$$1 - (2n-1) 2^{2n} B_{2n} + (2n-1) (2-2^{2n}) B_{2n} = 1 + (2n-1) (2-2^{2n+1}) B_{2n} \\ = 1 + 2 (2n-1) (1-2^{2n}) B_{2n}.$$
Our formula now reads
$$B_{2n} = \frac{1}{2(2n-1)(2^{2n}-1)} \left(1 - 1 - 2 (2n-1) (1-2^{2n}) B_{2n}\right)$$
and we have evidently proved the claim.
Addendum. The simplest way of seeing that the residues of the main function sum to zero is to use the square $\pm \pi(2N+1) \pm \pi(2N+1)i$ for a contour in the evaluation that passes midway between the poles on the imaginary axis. It then becomes possible to apply ML and show that the integral vanishes in $N,$ substantiating the claim. Number the sides $\Gamma_{0,1,2,3}$ starting with the right side. The length of every one of the four sides is $2\pi(2N+1).$ We get for $\Gamma_0$ where we parameterize with $t$ through $z=\pi(2N+1)+it$ where $-\pi(2N+1)\le t\le \pi(2N+1)$
$$\left| g(z) \right| = \frac{1}{|\pi(2N+1)+it|^{2n-1}} \left|\frac{\exp(3(\pi(2N+1)+it)/2)}{(\exp(\pi(2N+1)+it)-1)^2}\right| \\ \le \frac{1}{\pi^{2n-1} (2N+1)^{2n-1}} \frac{\exp(3\pi(2N+1)/2)} {(\exp(\pi(2N+1))\cos t - 1)^2 + (\exp(\pi(2N+1)) \sin t)^2} \\ = \frac{1}{\pi^{2n-1}(2N+1)^{2n-1}} \frac{\exp(3\pi(2N+1)/2)} {\exp(2\pi(2N+1)) - 2 \exp(\pi(2N+1)) \cos t + 1} \\ = \frac{1}{\pi^{2n-1}(2N+1)^{2n-1}} \frac{\exp(-\pi(2N+1)/2)} {1 - 2 \exp(-\pi(2N+1)) \cos t + \exp(-2\pi(2N+1))}.$$
Here we have used the fact that the point on $\Gamma_0$ closest to the origin is the intersection with the real axis to bound $|1/z^{2n-1}|$. Now clearly the terms in the denominator that are not constant both vanish (the cosine has $|\cos t|\le 1$) and we get using ML for $N$ large enough the upper bound
$$\frac{2\pi(2N+1)}{\pi^{2n-1}(2N+1)^{2n-1}} \frac{\exp(-\pi(2N+1)/2)}{1/2} = \frac{4\exp(-\pi(2N+1)/2)}{\pi^{2n-2}(2N+1)^{2n-2}}$$
so that the contribution from $\Gamma_0$ vanishes in the limit. Next we get for $\Gamma_2$ where we parameterize with $t$ through $z=-\pi(2N+1)+it$ where $-\pi(2N+1)\le t\le \pi(2N+1)$
$$\frac{1}{\pi^{2n-1}(2N+1)^{2n-1}} \frac{\exp(-3\pi(2N+1)/2)} {\exp(-2\pi(2N+1)) - 2 \exp(-\pi(2N+1)) \cos t + 1}.$$
We once more get two vanishing terms in the denominator and the ML bound
$$\frac{4\exp(-3\pi(2N+1)/2)}{\pi^{2n-2}(2N+1)^{2n-2}}$$
so this vanishes too. To conclude we need the two horizontal segments $\Gamma_1$ and $\Gamma_3$ which we parameterize with $z = t \pm\pi(2N+1)i$ where $-\pi(2N+1)\le t\le \pi(2N+1).$ We find
$$\left| g(z) \right| = \frac{1}{|t\pm\pi(2N+1)i|^{2n-1}} \left|\frac{\exp(3(t\pm\pi(2N+1)i)/2)}{(\exp(t\pm\pi(2N+1)i)-1)^2}\right| \\ \le \frac{1}{\pi^{2n-1}(2N+1)^{2n-1}} \frac{\exp(3t/2)} {(\exp t \cos(\pm\pi(2N+1))-1)^2 + (\exp t \sin(\pm\pi(2N+1)))^2} \\ = \frac{1}{\pi^{2n-1}(2N+1)^{2n-1}} \frac{\exp(3t/2)} {\exp (2t) - 2 \exp t\cos(\pm\pi(2N+1)) + 1} \\ = \frac{1}{\pi^{2n-1}(2N+1)^{2n-1}} \frac{\exp(3t/2)}{(\exp t +1)^2}.$$
We see there are no singularities in $t$ and that for $t$ going to minus infinity this is asymptotic to $\exp(3t/2)$ with the real axis as asymptote from above and with $t$ going to plus infinity it is asymptotic to $\exp(-t/2)$, with the same asymptote, also from above. With $t$ real we differentiate to obtain:
$$\frac{1}{2} \frac{\exp(3t/2) (3 - \exp t)}{(\exp t +1)^3}$$
By the sign of this derivative the function is increasing to a maximum at $t=\log 3$ and decreasing thereafter. Putting it all together the value at $t=\log 3$ which is $\frac{1}{16} 3^{3/2}$ serves as a constant upper bound independent of $N.$ The ML bound becomes
$$\frac{1}{8} 3^{3/2} \frac{1}{\pi^{2n-2}(2N+1)^{2n-2}}$$
which also vanishes in $N$ (we have used $n\ge 2$ throughout). This shows that the integral along the square vanishes as $N$ goes to infinity and hence the residues at the poles inside the circle must indeed sum to zero as claimed and we have justified the construction used above.