Identity related to $\sum_{k=0}^{n}\frac{x^k}{\binom{n}{k}}$

Question

How it can be shown that:

$$\sum_{k=0}^{n}\frac{x^k}{\binom{n}{k}}=\left(n+1\right)\left(\frac{x}{x+1}\right)^{n+1}\sum_{k=1}^{n+1}\frac{1+x^k}{\left(1+x\right)k}\left(\frac{1+x}{x}\right)^{k}$$

for $x \ne-1$

I tried Additive Forms of Reciprocal Pascal’s Identity , but could not derive that.

Answer 1

Preliminaries

Note that $$ \begin{align} \sum_{k=1}^\infty\sum_{n=0}^\infty\frac{(-1)^k}k\binom{m}{n+1-k}x^n &=\sum_{k=1}^\infty\sum_{n=k-1}^\infty\frac{(-1)^k}k\binom{m}{n+1-k}x^n\tag{1a}\\ &=\sum_{k=1}^\infty\sum_{n=0}^\infty\frac{(-1)^k}k\binom{m}{n}x^{n+k-1}\tag{1b}\\ &=\sum_{k=1}^\infty\frac{(-1)^k}kx^{k-1}(1+x)^m\tag{1c}\\ &=-\frac{(1+x)^m}x\,\log(1+x)\tag{1d} \end{align} $$ Explanation:
$\text{(1a)}$: $\binom{m}{n+1-k}=0$ for $n\lt k-1$
$\text{(1b)}$: substitute $n\mapsto n+k-1$
$\text{(1c)}$: Binomial Theorem
$\text{(1d)}$: apply the series for $\log(1+x)$

and $$ \begin{align} \sum_{k=1}^\infty\sum_{n=0}^\infty\frac1k\binom{m-k}{n+1-k}x^n &=\sum_{k=1}^\infty\sum_{n=k-1}^\infty\frac1k\binom{m-k}{n+1-k}x^n\tag{2a}\\ &=\sum_{k=1}^\infty\sum_{n=0}^\infty\frac1k\binom{m-k}{n}x^{n+k-1}\tag{2b}\\ &=\sum_{k=1}^\infty\frac1kx^{k-1}(1+x)^{m-k}\tag{2c}\\ &=-\frac{(1+x)^m}x\log\left(1-\frac{x}{1+x}\right)\tag{2d}\\[6pt] &=\frac{(1+x)^m}x\log(1+x)\tag{2e} \end{align} $$ Explanation:
$\text{(2a)}$: $\binom{m}{n+1-k}=0$ for $n\lt k-1$
$\text{(2b)}$: substitute $n\mapsto n+k-1$
$\text{(2c)}$: Binomial Theorem
$\text{(2d)}$: apply the series for $\log(1-x)$
$\text{(2e)}$: simplify

Adding $(1)$ and $(2)$ and examining coefficients of $x^n$ gives $$ \sum_{k=1}^{n+1}\left[\frac{(-1)^k}k\binom{m}{n+1-k}+\frac1k\binom{m-k}{n+1-k}\right]=0\tag3 $$ Lemma $\bf{1}$: $$ \sum_{k=0}^m\frac{(-1)^{m-k}}{n+1-k}\binom{m}{k}=\frac1{(n+1)\binom{n}{m}}\tag4 $$ Proof: Case $m=0$ is trivial. Suppose it is true for some $m$. $$ \begin{align} \sum_{k=0}^{m+1}\frac{(-1)^{m-k+1}}{n+1-k}\binom{m+1}{k} &=\sum_{k=0}^{m+1}\frac{(-1)^{m-k+1}}{n+1-k}\left[\binom{m}{k-1}+\binom{m}{k}\right]\tag{4a}\\ &=\sum_{k=0}^m\frac{(-1)^{m-k}}{n-k}\binom{m}{k}-\sum_{k=0}^m\frac{(-1)^{m-k}}{n+1-k}\binom{m}{k}\tag{4b}\\ &=\frac1{n\binom{n-1}{m}}-\frac1{(n+1)\binom{n}{m}}\tag{4c}\\ &=\frac1{(m+1)\binom{n}{m+1}}-\frac1{(m+1)\binom{n+1}{m+1}}\tag{4d}\\ &=\frac{\binom{n}{m}}{(m+1)\binom{n}{m+1}\binom{n+1}{m+1}}\tag{4e}\\[3pt] &=\frac1{(n+1)\binom{n}{m+1}}\tag{4f} \end{align} $$ Explanation:
$\text{(4a)}$: Pascal triangle recursion
$\text{(4b)}$: substitute $k\mapsto k+1$ in the left sum
$\text{(4c)}$: apply the Lemma for $m$
$\text{(4d)}$: $(n+1)\binom{n}{m}=(m+1)\binom{n+1}{m+1}$
$\text{(4e)}$: $\binom{n+1}{m+1}-\binom{n}{m+1}=\binom{n}{m}$
$\text{(4f)}$: $(m+1)\binom{n+1}{m+1}=(n+1)\binom{n}{m}$

$\large\square$

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Computing the Coefficients of the Series

Start with the series with the more complicated terms $$ \begin{align} &(n+1)\left(\frac{x}{x+1}\right)^{n+1}\sum_{k=1}^{n+1}\frac{1+x^k}{(1+x)k}\left(\frac{1+x}x\right)^k\\ &=(n+1)\sum_{k=1}^{n+1}\frac{1+x^k}{(1+x)k}\left(\frac{x}{1+x}\right)^{n+1-k}\tag{5a}\\ &=(n+1)\sum_{k=1}^{n+1}\frac1k\frac{x^{n+1-k}+x^{n+1}}{(1+x)^{n+2-k}}\tag{5b}\\ &=(n+1)\sum_{k=1}^{n+1}\sum_{j=0}^\infty\frac1k\binom{-n-2+k}{j}\left(x^{n+1-k+j}+x^{n+1+j}\right)\tag{5c}\\ &=(n+1)\sum_{k=1}^{n+1}\sum_{j=0}^\infty\frac{(-1)^j}k\binom{n+1-k+j}{j}\left(x^{n+1-k+j}+x^{n+1+j}\right)\tag{5d}\\ &=(n+1)\sum_{k=1}^{n+1}\sum_{m=n+1-k}^\infty\frac{(-1)^{m+k-n-1}}k\binom{m}{m+k-n-1}x^m\\ &+(n+1)\sum_{k=1}^{n+1}\sum_{m=n+1}^\infty\frac{(-1)^{m-n-1}}k\binom{m-k}{m-n-1}x^m\tag{5e}\\ &=(n+1)\sum_{k=1}^{n+1}\sum_{m=n+1-k}^\infty\frac{(-1)^{m+k-n-1}}k\binom{m}{n+1-k}x^m\\ &+(n+1)\sum_{k=1}^{n+1}\sum_{m=n+1}^\infty\frac{(-1)^{m-n-1}}k\binom{m-k}{n+1-k}x^m\tag{5f}\\ &=(n+1)\sum_{k=1}^{n+1}\sum_{m=n+1-k}^n\frac{(-1)^{m+k-n-1}}k\binom{m}{n+1-k}x^m\tag{5g}\\ &=(n+1)\sum_{m=0}^n\sum_{k=n+1-m}^{n+1}\frac{(-1)^{m+k-n-1}}k\binom{m}{n+1-k}x^m\tag{5h}\\ &=(n+1)\sum_{m=0}^n\sum_{k=0}^m\frac{(-1)^{m-k}}{n+1-k}\binom{m}{k}x^m\tag{5i}\\ &=(n+1)\sum_{m=0}^n\frac{x^m}{(n+1)\binom{n}{m}}\tag{5j}\\ &=\sum_{m=0}^n\frac{x^m}{\binom{n}{m}}\tag{5k} \end{align} $$ Explanation:
$\text{(5a)}$: distribute the factor out front over all terms
$\text{(5b)}$: move factors about
$\text{(5c)}$: expand the denominator with the Binomial Theorem
$\text{(5d)}$: negative binomial coefficients
$\text{(5e)}$: $j\mapsto m+k-n-1$ in left sum and $j\mapsto m-n-1$ in right sum
$\text{(5f)}$: $\binom{n}{k}=\binom{n}{n-k}$ when $n$ is a non-negative integer
$\text{(5g)}$: apply $(3)$ for $m\ge n+1$
$\text{(5h)}$: swap order of summation
$\text{(5i)}$: substitute $k\mapsto n+1-k$
$\text{(5j)}$: Lemma $1$
$\text{(5k)}$: simplify

Answer 2

Your result is a kind of generalization of Newton's binomial identity with binomial coefficients replaced by their inverses.

You will find in page 2 of this reference by Toufik Mansour, University of Haifa) the more general expression :

$$\sum_{k=0}^{n}\frac{a^kb^{n-k}}{\binom{n}{k}}=\frac{n+1}{(a+b)\left(\tfrac{1}{a}+\tfrac{1}{b}\right)^{n+1}}\sum_{k=1}^{n+1}\frac{(a^k+b^k)\left(\tfrac{1}{a}+\tfrac{1}{b}\right)^{k}}{k}\tag{1}$$

with its proof. Nice expression...

It suffices now to take $a=x$ and $b=1$...

I have to admit that the proof is long... and uses generating functions.

Remark : If in the following identity for Beta integrals (see here):

$$B(x,y)=\int_{t=0}^{t=1}t^{x-1}(1-t)^{y-1}dt=\frac{(x-1)!(y-1)!}{(x+y-1)!}\tag{2}$$

one takes $x-1=k$ and $y-1=n-k$, we deduce that :

$$\frac{1}{\binom{n}{k}}=(n+1)\int_{t=0}^{t=1}t^k(1-t)^{n-k}dt\tag{3}$$

(besides, (3) is mentionned in the referenced article).

It is likely that a (simpler) proof of your identity can be deduced from (3) by multiplying it by $x^k$ and summing from $k=0$ to $k=n$.