# How can we show that $\sum_{k=1}^{n}{2n+2\choose 2k}B_{2k}=n?$

Given the sum

$$\sum_{k=1}^{n}{2n+2\choose 2k}B_{2k}=n\tag1$$ Where $B_{2k}$ is Bernoulli number

It is quite interesting to me, the answer results in a natural number, how do you go about showing that? I haven't got any ideas where to start.

• You could look at the paper of Štofka referenced here, as well as the related papers of Lehmer and Carlitz: those are similar-looking results that might give you some ideas or even a reference. – Greg Martin Mar 18 '17 at 6:47
• I suspect generating functions would make this possible. – Patrick Stevens Mar 18 '17 at 23:25

We seek to evaluate

$$Q_n = \sum_{k=1}^n {2n+2\choose 2k} B_{2k}.$$

With this in mind we introduce the generating function

$$G(z) = \sum_{n\ge 1} Q_n \frac{z^{2n+2}}{(2n+2)!}.$$

We then obtain for $G(z)$ that

$$G(z) = \sum_{n\ge 1} \frac{z^{2n+2}}{(2n+2)!} \sum_{k=1}^n {2n+2\choose 2k} B_{2k} \\ = \sum_{n\ge 1} \frac{z^{2n+2}}{(2n+2)!} \sum_{k=1}^n {2n+2\choose 2k} (2k)! [w^{2k}] \frac{w}{\exp(w)-1} \\ = \sum_{n\ge 1} z^{2n+2} \sum_{k=1}^n \frac{1}{(2n+2-2k)!} [w^{2k}] \frac{w}{\exp(w)-1} \\ = \sum_{k\ge 1} [w^{2k}] \frac{w}{\exp(w)-1} \sum_{n\ge k} z^{2n+2} \frac{1}{(2n+2-2k)!} \\ = \sum_{k\ge 1} z^{2k} [w^{2k}] \frac{w}{\exp(w)-1} \sum_{n\ge 0} z^{2n+2} \frac{1}{(2n+2)!} \\ = \left(-1 + \frac{1}{2}(\exp(z)+\exp(-z))\right) \sum_{k\ge 1} z^{2k} [w^{2k}] \frac{w}{\exp(w)-1} \\ = \left(-1 + \frac{1}{2}(\exp(z)+\exp(-z))\right) \left(-1 + \frac{1}{2} z + \frac{z}{\exp(z)-1}\right).$$

As we examine this formula we see that it could have been obtained by inspection. We put $\exp(z) = v$ while we simplify and get

$$\frac{1}{v-1} \left(-1 + \frac{1}{2}v+\frac{1}{2}\frac{1}{v}\right) \left(1-v + \frac{1}{2}zv - \frac{1}{2}z+z\right) \\ = \frac{1}{2}\frac{1}{v}\frac{1}{v-1} (-2v+v^2+1)\left(1-v+\frac{1}{2}zv+\frac{1}{2}z\right) \\ = \frac{1}{2}\frac{1}{v} (v-1) \left(1-v+\frac{1}{2}zv+\frac{1}{2}z\right) \\ = \frac{1}{2}\frac{1}{v} \left(-1+v-\frac{1}{2}zv-\frac{1}{2}z + v-v^2+\frac{1}{2}zv^2+\frac{1}{2}zv\right) \\ = \frac{1}{2}\frac{1}{v} \left(-1+2v-\frac{1}{2}z -v^2+\frac{1}{2}zv^2\right) \\ = \frac{1}{2} \left(-\frac{1}{v}+2-\frac{1}{2}z\frac{1}{v} -v+\frac{1}{2}zv\right)$$

Extracting coefficients from this now yields

$$(2n+2)! [z^{2n+2}] \frac{1}{2} \left(-\frac{1}{v}+2-\frac{1}{2}z\frac{1}{v} -v+\frac{1}{2}zv\right) \\ = \frac{1}{2} \left(-(-1)^{2n+2} - \frac{1}{2} (2n+2) (-1)^{2n+1} - 1 + \frac{1}{2} (2n+2) \right) \\ = \frac{1}{2} \left(-2 + (n+1) + (n+1)\right).$$

We thus have

$$\bbox[5px,border:2px solid #00A000]{Q_n = n.}$$

Observe that we may verify that the coeffcients of odd powers are zero. We obtain

$$(2n+1)! [z^{2n+1}] \frac{1}{2} \left(-\frac{1}{v}+2-\frac{1}{2}z\frac{1}{v} -v+\frac{1}{2}zv\right) \\ = \frac{1}{2} \left(-(-1)^{2n+1} - \frac{1}{2} (2n+1) (-1)^{2n} - 1 + \frac{1}{2} (2n+1)\right) \\ = \frac{1}{2} \left(- \frac{1}{2} (2n+1) + \frac{1}{2} (2n+1)\right) = 0.$$

The same kind of arithmetic works for $[z^0].$

Remark. The above used the technique of annihilated coefficient extractors (ACE), otherwise known as the substitution rule for formal power series. There are several more examples at this MSE link I and at this MSE link II and also here at this MSE link III.