How to prove that $\sum_{i=1}^{m} 2^{i} B_{i}\binom{m}{i} \frac{1}{m-i+1}=\frac{2 m+1}{m+1}$ when $m$ is an even positive integer?

Question

I want to prove the identity below $$ \sum_{i=1}^{m} 2^{i} B_{i}\binom{m}{i} \frac{1}{m-i+1}=\frac{2 m+1}{m+1} $$ I've carried out some computations and verified it but it only seems to hold when $m$ is an even number. Does someone have a proof of this? Preferably without induction? Maybe multiplying $\frac{z}{1-e^{-z}}$ by some $f(z)(e^z-1)$ and then comparing coefficients of $z^n$? I tried starting off with $$\frac{2z}{1-e^{-2z}} = \sum_{k=0}^{\infty} \frac{2^kz^k}{k!} B_k^+ $$ but I didn't end up anywhere useful as I just got $$ \sum_{m=0}^{\infty} \frac{2^{m+1} z^{m+1}}{m !}=\sum_{m=0}^{\infty} z^{m+1} \sum_{k=0}^{m} \frac{2^{m+1} B_{k}^{+}}{k !(m-k+1) !} $$ and comparing coefficients just gave the classic identity $$ \sum_{m=0}^{k} B_m^+ \binom{m}{k}\frac{1}{m-k+1} = 1 $$ This does look similar to $$ \sum_{i=1}^{m} 2^{i} B_{i}\left(\begin{array}{c} m \\ i \end{array}\right) \frac{1}{m-i+1}=\frac{2 m+1}{m+1} $$ Or rather, since $m$ is even (so hence $m = 2k$ for some $ k \in \mathbb{N}$) $$ \sum_{i=1}^{2k} 2^{i} B_{i}\left(\begin{array}{c} 2k \\ i \end{array}\right) \frac{1}{2k-i+1}=\frac{4k+1}{2k+1} $$ Any help would be greatly appreciated, I tried using induction but $\binom{2k+2}{i}$ causes problems and I can't prove the inductive step using my induction hypothesis. Also, I wonder why the formula only holds for even $m$. Is there a related one for odd $m$?

Answer 1

Here is a simple proof that works when $ n $ is even :

Let $ n\in 2\mathbb{N}^{*} :$

\begin{aligned} \sum_{k=0}^{n}{B_{k}\binom{n}{k}\frac{2^{k}}{n-k+1}}&=2^{n°1}\sum_{k=0}^{n}{B_{k}\binom{n}{k}\int_{0}^{\frac{1}{2}}{x^{n-k}\,\mathrm{d}x}}\\&=2^{n+1}\int_{0}^{\frac{1}{2}}{B_{n}\left(x\right)\mathrm{d}x} \end{aligned}

Using the substitution $ \left\lbrace\begin{aligned}y&=1-x\\ \mathrm{d}y &=-\,\mathrm{d}x\end{aligned}\right.$, we get : $$ \int_{0}^{\frac{1}{2}}{B_{n}\left(x\right)\mathrm{d}x}=\int_{\frac{1}{2}}^{1}{B_{n}\left(1-x\right)\mathrm{d}x}=\left(-1\right)^{n}\int_{\frac{1}{2}}^{1}{B_{n}\left(x\right)\mathrm{d}x}=\int_{\frac{1}{2}}^{1}{B_{n}\left(x\right)\mathrm{d}x} $$

Thus : $$ \int_{0}^{\frac{1}{2}}{B_{n}\left(x\right)\mathrm{d}x}=\frac{1}{2}\int_{0}^{1}{B_{n}\left(x\right)\mathrm{d}x} = 0 $$

That means $$ \sum_{k=0}^{n}{B_{k}\binom{n}{k}\frac{2^{k}}{n-k+1}}=0 $$

Now let's try to get the general result using a different method.

Denoting $ \left(\forall n\in\mathbb{N}\right),\ u_{n}=\sum\limits_{k=0}^{n}{B_{k}\binom{n}{k}\frac{2^{k}}{n-k+1}} $ we can observe that if $ x $ is such that $ \left|x\right|<2\pi $, $ \sum\limits_{n\geq 0}{\frac{B_{n}}{n!}x^{n}} $ and $ \sum\limits_{n\geq 0}{\frac{2^{-n}}{\left(n+1\right)!}x^{n}} $ both converge. Thus the series $ \sum\limits_{n\geq 0}{\frac{u_{n}}{2^{n}n!}x^{n}} $ also converges and we have : \begin{aligned} \sum_{n=0}^{+\infty}{\frac{u_{n}}{2^{n}n!}x^{n}}&=\left(\sum_{n=0}^{+\infty}{\frac{B_{n}}{n!}x^{n}}\right)\left(\sum_{n=0}^{+\infty}{\frac{2^{-n}}{\left(n+1\right)!}x^{n}}\right)\\ &=\frac{x}{\mathrm{e}^{x}-1}\times\frac{\mathrm{e}^{\frac{x}{2}}-1}{\frac{x}{2}}\\ &=\frac{1}{\sinh{\left(\frac{x}{2}\right)}}-\frac{2}{\mathrm{e}^{x}-1}\\ &=\frac{2}{x}-\sum_{n=1}^{+\infty}{B_{2n}\frac{2\left(1-2^{1-2n}\right)}{\left(2n\right)!}x^{2n-1}}-\frac{2}{x}\sum_{n=0}^{+\infty}{\frac{B_{n}}{n!}x^{n}}\\ &=1-\sum_{n=1}^{+\infty}{B_{2n}\frac{4\left(1-2^{-2n}\right)}{\left(2n\right)!}x^{2n-1}}\end{aligned}

From that formula we get the following : $$\fbox{$\begin{array}{rcl}\begin{aligned} u_{0} &=1\\ \left(\forall n\geq 1\right),\ u_{2n} &= 0\\ \left(\forall n\geq 1\right),\ u_{2n-1}&=B_{2n}\frac{1-2^{2n}}{n} \end{aligned} \end{array}$}$$

From these expressions you can get expressions for the sum $ \sum\limits_{k=1}^{n}{B_{k}^{+}\binom{n}{k}\frac{2^{k}}{n-k+1}} $ : $$ \fbox{$\begin{array}{rcl}\left(\forall n\in\mathbb{N}^{*}\right),\ \displaystyle\sum_{k=1}^{n}{B_{k}^{+}\binom{n}{k}\frac{2^{k}}{n-k+1}}=\frac{2n+1}{n+1} + u_{n} \end{array}$}$$

Answer 2

We seek to show that for $m$ even

$$\sum_{q=1}^m 2^q B^+_q {m\choose q} \frac{1}{m-q+1} = \frac{2m+1}{m+1}$$

or alternatively for $m$ even

$$\sum_{q=1}^m 2^q B^+_q {m+1\choose q} = 2m+1$$

where

$$B^+_q = q! [z^q] \frac{z}{1-\exp(-z)}.$$

We get for the LHS

$$-B^+_0 - 2^{m+1} B^+_{m+1} + \sum_{q=0}^{m+1} 2^q B^+_q {m+1\choose q} \\ = - 1 - 2^{m+1} B^+_{m+1} + (m+1)! [z^{m+1}] \exp(z) \frac{2z}{1-\exp(-2z)} \\ = - 1 - (m+1)! [z^{m+1}] \frac{2z}{1-\exp(-2z)} \\ + (m+1)! [z^{m+1}] \exp(z) \frac{2z}{1-\exp(-2z)} \\ = - 1 + (m+1)! [z^{m+1}] \frac{2z (\exp(z)-1)}{1-\exp(-z)^2} \\ = - 1 + (m+1)! [z^{m+1}] \frac{2z \exp(z) (1-\exp(-z))}{1-\exp(-z)^2} \\ = - 1 + (m+1)! [z^{m+1}] \frac{2z \exp(z)}{1+\exp(-z)} \\ = - 1 + (m+1)! [z^{m+1}] \left(\frac{2z \exp(z) + 2z}{1+\exp(-z)} - \frac{2z}{1+\exp(-z)} \right) \\ = - 1 + (m+1)! [z^{m+1}] \left(2z \exp(z) - \frac{2z}{1+\exp(-z)} \right) \\ = - 1 + 2 (m+1)! [z^{m}] \left(\exp(z) - \frac{1}{1+\exp(-z)} \right) \\ = - 1 + 2 (m+1)! \frac{1}{m!} - 2 (m+1)! [z^{m}] \frac{1}{1+\exp(-z)} \\ = 2m+1 - 2 (m+1)! [z^{m}] \frac{1}{1+\exp(-z)}.$$

We observe that this evaluates to zero when $m=0$ as claimed and henceforth assume $m\ge 1.$ Continuing,

$$2m+1 - 2 (m+1)! [z^{m}] \left( \frac{1/2-\exp(-z)/2}{1+\exp(-z)} + \frac{1}{2}\right).$$

Now note that with $f(z) = \frac{1/2-\exp(-z)/2}{1+\exp(-z)}$ we have

$$f(-z) = \frac{1/2-\exp(z)/2}{1+\exp(z)} = \frac{\exp(-z)/2-1/2}{\exp(-z)+1} = - f(z)$$

so the function $f(z)$ is odd and hence the even order coefficients of its Taylor series vanish. Since the constant $1/2$ does not contribute to $[z^m]$ when $m \ge 1$ that leaves just $2m+1$ and we have the desired result. For the odd order coefficients we have to evaluate the remaining coefficient extractor.

$$2 (m+1)! [z^{m}] \frac{1}{1+\exp(-z)} = 2 (m+1)! (-1)^m [z^{m}] \frac{1}{1+\exp(z)} \\ = 2 (m+1)! (-1)^m [z^{m}] \frac{1}{2+\exp(z)-1} \\ = (m+1)! (-1)^m [z^{m}] \frac{1}{1+(\exp(z)-1)/2} \\ = (m+1)! (-1)^m [z^{m}] \sum_{q=0}^m (-1)^q 2^{-q} (\exp(z)-1)^q \\ = (m+1) (-1)^m \sum_{q=0}^m (-1)^q 2^{-q} q! {m\brace q}.$$

We thus obtain the closed form

$$\bbox[5px,border:2px solid #00A000]{ 2m+1 - [[m \;\text{odd}]] (m+1) (-1)^m \sum_{q=1}^m (-1)^q 2^{-q} q! {m\brace q}.}$$

The Iverson bracket is not essential here as the sum evaluates to zero anyway when $m$ is even. This yields for the original as proposed by OP that it is

$$\bbox[5px,border:2px solid #00A000]{ \frac{2m+1}{m+1} - [[m \;\text{odd}]] (-1)^m \sum_{q=1}^m (-1)^q 2^{-q} q! {m\brace q}.}$$