Summation $\sum_{k=0}^{\lfloor\frac{n-1}{2}\rfloor}\frac{1}{\alpha^{2k+1}}\frac{B_{2k+2}}{\left(n-2k\right)!\left(2k+2\right)!}$

Question

For $\alpha, n \in \text{N}$
How do we Prove That,
$$\sum_{k=0}^{\lfloor\frac{n-1}{2}\rfloor}\frac{1}{\alpha^{2k+1}}\frac{B_{2k+2}}{\left(n-2k\right)!\left(2k+2\right)!}=\frac{\alpha}{n!\left(n+2\right)}-\frac{2\alpha-1}{2\left(n+1\right)!}+\frac{1}{\left(n+1\right)!\alpha^{n+1}}\sum_{j=0}^{\alpha-1}j^{n+1}$$ Where, $B_{n}$ is the $n$th Bernoulli Number.
The right side of the equation does not contain Bernoulli Numbers, I am guessing this used a property but I am not aware of that.

Answer 1

I'll be using some conventions and formulae present on the Wikipedia page for the Bernoulli numbers at the section the link goes to, but for self-containment (and also to adapt them to the current notations of this question) I'll write them down here as well:
Let $(B^{+}_n)_n$ et $(B^{-}_n)_n$ be the two conventions for the Bernoulli numbers, only differing at $n = 1$, where $B^{\pm}_1 = \pm \frac{1}{2}$. Since this'll play a role as well during the calculation, I'll also state the value of $B_0 = 1$.
Then, we'll consider as known the following formulae, aptly named Bernoulli's formulae: $$(n+2)\sum_{j = 0}^\alpha j^{n+1} = \sum_{k = 0}^{n+1} \binom{n+2}{k} B^+_k \alpha^{n+2-k} = \sum_{k = 0}^{n+1} \binom{n+2}{k} B^-_k (-1)^k \alpha^{n+2-k}$$

Let's denote by $S_n$ the LHS of OP's expression. We finally have, multiplying it by $(n+2)!\alpha^{n+1}$ to introduce the missing binomial coefficients and correct the powers on $\alpha$ to use said formulae: $$\begin{split} (n+2)!\alpha^{n+1} S_n &= (n+2)!\alpha^{n+1} \sum_{k=0}^{\lfloor\frac{n-1}{2}\rfloor}\frac{1}{\alpha^{2k+1}}\frac{B_{2k+2}}{\left(n-2k\right)!\left(2k+2\right)!}\\ &= \sum_{k=0}^{\lfloor\frac{n-1}{2}\rfloor} \alpha^{n+1-(2k+1)}B_{2k+2}\binom{n+2}{2k+2}\\ &= \sum_{k=1}^{\lfloor\frac{n+1}{2}\rfloor} \alpha^{n+2-2k}B_{2k}\binom{n+2}{2k}\\ &= \sum_{k=2 \atop k \text{ even}}^{n+1} \alpha^{n+2-k}B_{k}\binom{n+2}{k}\\ &= \sum_{k=2}^{n+1} \frac{1 + (-1)^k}{2}\alpha^{n+2-k}B_{k}\binom{n+2}{k}\\ &= \frac{1}{2}\Bigg(\sum_{k = 2}^{n+1} \binom{n+2}{k} B^+_k \alpha^{n+2-k} + \sum_{k = 2}^{n+1} \binom{n+2}{k} B^-_k (-1)^k \alpha^{n+2-k}\Bigg)\\ &= \frac{1}{2}\Bigg(\sum_{k = 0}^{n+1} \binom{n+2}{k} B^+_k \alpha^{n+2-k} + \sum_{k = 0}^{n+1} \binom{n+2}{k} B^-_k (-1)^k \alpha^{n+2-k}\\ &\quad\quad\quad - \binom{n+2}{0}B_0^+\alpha^{n+2-0} - \binom{n+2}{0}B_0^-(-1)^0\alpha^{n+2-0}\\ &\quad\quad\quad - \binom{n+2}{1}B_1^+\alpha^{n+2-1} - \binom{n+2}{1}B_1^-(-1)^1\alpha^{n+2-1}\Bigg)\\ &= (n+2)\sum_{j=0}^{\alpha} j^{n+1} - \alpha^{n+2} - \frac{1}{2}(n+2)\alpha^{n+1}\\ &= (n+2)\sum_{j=0}^{\alpha - 1} j^{n+1} - \alpha^{n+2} + \frac{1}{2}(n+2)\alpha^{n+1}\end{split}$$ Redividing by $(n+2)! \alpha^{n+1}$, this then grants: $$S_n = \frac{1}{(n+1)!\alpha^{n+1}}\sum_{j=0}^{\alpha - 1} j^{n+1} - \frac{\alpha}{(n+2)!} + \frac{1}{2(n+1)!}$$ We conclude by noticing that: $$\frac{1}{(n+2)!} = \frac{1}{(n+2)(n+1)n!} = \frac{1}{n!}\left(\frac{1}{n+1} - \frac{1}{n+2}\right) = \frac{1}{(n+1)!} - \frac{1}{n!(n+2)}$$ which implies that: $$\begin{split}S_n &= \frac{1}{(n+1)!\alpha^{n+1}}\sum_{j=0}^{\alpha - 1} j^{n+1} - \left(\frac{\alpha}{(n+1)!} - \frac{\alpha}{n!(n+2)}\right) + \frac{1}{2(n+1)!}\\ &= \frac{1}{(n+1)!\alpha^{n+1}}\sum_{j=0}^{\alpha - 1} j^{n+1} + \frac{\alpha}{n!(n+2)} - \frac{\alpha}{(n+1)!} + \frac{1}{2(n+1)!}\\ &= \frac{1}{(n+1)!\alpha^{n+1}}\sum_{j=0}^{\alpha - 1} j^{n+1} + \frac{\alpha}{n!(n+2)} - \frac{2\alpha - 1}{2(n+1)!}\end{split}$$ which is what was desired.

Answer 2

We first try to evaluate

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

and we note that this is

$$\sum_{k=0}^n \frac{1}{\alpha^{k+1}} \frac{B_{k+2}}{(n-k)! (k+2)!}$$

We get with the EGF of the Bernoulli numbers

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

Here we have extended to infinity due to the coefficient extractor. Continuing,

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

Now observe that

$$\sum_{j=0}^{\alpha-1} j^{n+1} = (n+1)! \sum_{j=0}^{\alpha-1} [w^{n+1}] \exp(jw) \\ = (n+1)! [w^{n+1}] \frac{\exp(w\alpha)-1}{\exp(w)-1} \\ = (n+1)! [w^{n+2}] \frac{w\exp(w\alpha)-w}{\exp(w)-1} \\ = - \frac{B_{n+2}}{n+2} + (n+1)! [w^{n+2}] \frac{w\exp(w\alpha)}{\exp(w)-1} \\ = - \frac{B_{n+2}}{n+2} + \alpha^{n+1} (n+1)! [w^{n+2}] \frac{w\exp(w)}{\exp(w/\alpha)-1}.$$

Merge the two to get

$$- \frac{\alpha}{(n+2)!} + \frac{1}{2} \frac{1}{(n+1)!} + \frac{1}{\alpha^{n+1} (n+1)!} \sum_{j=0}^{\alpha-1} j^{n+1} + \frac{B_{n+2}}{\alpha^{n+1} (n+2)!}.$$

To conclude recall that OP asks for the upper limit $\lfloor (n-1)/2\rfloor$ rather than $\lfloor n/2 \rfloor.$ These are equal when $n$ is odd but when $n$ is even the top term is missing. This term is (substitute and simplify)

$$\frac{1}{\alpha^{n+1}} \frac{B_{n+2}}{(n+2)!}$$

so that we have

$$- \frac{\alpha}{(n+2)!} + \frac{1}{2} \frac{1}{(n+1)!} + \frac{1}{\alpha^{n+1} (n+1)!} \sum_{j=0}^{\alpha-1} j^{n+1} \\ + \frac{B_{n+2}}{\alpha^{n+1} (n+2)!} (1-[[n\;\text{even}]]).$$

However when $n$ is odd we get zero from the Bernoulli number and when $n$ is even we get zero from the parenthesized term containing the Iverson bracket. We have shown that

$$\bbox[5px,border:2px solid #00A000]{ {\large \begin{array}{c|c} \sum_{k=0}^{\lfloor (n-1)/2 \rfloor} \frac{1}{\alpha^{2k+1}} \frac{B_{2k+2}}{(n-2k)! (2k+2)!} \\ = \frac{1}{2} \frac{n+2-2\alpha}{(n+2)!} + \frac{1}{\alpha^{n+1} (n+1)!} \sum_{j=0}^{\alpha-1} j^{n+1}. \end{array}}}$$

Remark. For the initial segment of the series about zero of $z/(\exp(z/\alpha)-1)$ we write

$$\alpha \frac{z/\alpha}{\exp(z/\alpha)-1} = \alpha \frac{B_0}{0!} + \alpha \frac{B_1}{1!} \frac{z}{\alpha} + \cdots = \alpha - \frac{1}{2} z + \cdots$$