Sum with Bernoulli poly. $\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n+1}{2k} B_{2k}(1/4) $

Question

I have a roundabout proof for $$ \sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n+1}{2k} B_{2k}(1/2)= \frac{n+1}{2^n} $$ where $B_{2k}(x)$ are the Bernoulli polynomials,$B_0(x)=1, B_2(x)=x^2-x+1/6,$ etc. Changing the argument of the Bernoulli polynomial and observing patterns (aided by OEIS A167205) I have a relationship which has been checked to high values of positive integer $n,$

$$ \sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n+1}{2k} B_{2k}(1/4)= \frac{n+1}{2^{2n \ - \ (1 \ - \ (-1)^n)/2)}} \ \frac{3^n+1}{3- \ (-1)^n }$$

How can this equation be proved?

Answer 1

We seek to show that with $B_{2k}(x)$ a Bernoulli polynomial we have

$$\sum_{k=0}^{\lfloor n/2 \rfloor} {n+1\choose 2k} B_{2k}(1/4) = \frac{1}{2^{2n+1}} (n+1) (1+3^n).$$

We will use the EGF of the Bernoulli polynomials which is

$$\sum_{n\ge 0} B_n(x) \frac{z^n}{n!} = \frac{z \exp(zx)}{\exp(z)-1}.$$

The LHS of the proposed identity is

$$(n+1) \sum_{k=0}^{\lfloor n/2 \rfloor} {n\choose 2k} B_{2k}(1/4) \frac{1}{n+1-2k}$$

so we really only have to prove that

$$\sum_{k=0}^{\lfloor n/2 \rfloor} {n\choose 2k} B_{2k}(1/4) \frac{1}{n+1-2k} = \frac{1}{2^{2n+1}} (1+3^n).$$

We get two pieces for the LHS, the first call it A is

$$\frac{1}{2} \sum_{k=0}^{n} {n\choose k} B_{k}(1/4) \frac{1}{n+1-k}$$

and the second call it B

$$\frac{1}{2} \sum_{k=0}^{n} {n\choose k} (-1)^k B_{k}(1/4) \frac{1}{n+1-k}.$$

What we have here is a convolution of two exponential generating functions. Recall that when we multiply two exponential generating functions of the sequences $\{p_n\}$ and $\{q_n\}$ we get that

$$ P(z) Q(z) = \sum_{n\ge 0} p_n \frac{z^n}{n!} \sum_{n\ge 0} q_n \frac{z^n}{n!} = \sum_{n\ge 0} \sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} p_k q_{n-k} z^n\\ = \sum_{n\ge 0} \sum_{k=0}^n \frac{n!}{k!(n-k)!} p_k q_{n-k} \frac{z^n}{n!} = \sum_{n\ge 0} \left(\sum_{k=0}^n {n\choose k} p_k q_{n-k}\right)\frac{z^n}{n!}$$

Observe that

$$\sum_{p\ge 0} \frac{1}{p+1} \frac{z^p}{p!} = \frac{\exp(z)-1}{z}.$$

Therefore the EGF of piece A is

$$\frac{1}{2} \exp(z/4)$$

and the EGF of piece B is

$$\frac{1}{2} \frac{-z \exp(-z/4)}{\exp(-z)-1} \frac{\exp(z)-1}{z} = \frac{1}{2} \frac{-z \exp(3z/4)}{1-\exp(z)} \frac{\exp(z)-1}{z} \\ = \frac{1}{2} \exp(3z/4).$$

Extracting the coefficient on $[z^n]$ (EGF not OGF) on A and B we find

$$\frac{1}{2} n! [z^n] (\exp(z/4) + \exp(3z/4)) = \frac{1}{2^{2n+1}} (1+3^n)$$

as required.