Showing this equality holds

We show the following identity for positive integers $a,b$ with $a-b\geq 1$ is valid: \begin{align*} \color{blue}{\sum_{k=1}^{a-b}\frac{(a-b-k)!}{(a+1-k)!}=\frac{1}{b\cdot b!}-\frac{(a-b)!}{b\cdot a!}}\tag{1} \end{align*}

We show (1) by writing the summands as reciprocal binomial coefficient $\binom{n}{k}^{-1}$. For non-negative values $n\geq k$ we have the nice integral representation \begin{align*} \binom{n}{k}^{-1}=(n+1)\int_{0}^1z^k(1-z)^{n-k}\,dz\tag{2} \end{align*}

We obtain \begin{align*} \color{blue}{\sum_{k=1}^{a-b}}&\color{blue}{\frac{(a-b-k)!}{(a+1-k)!}} =\sum_{k=1}^{a-b}\frac{(k-1)!}{(b+k)!}\tag{3.1}\\ &=\frac{1}{(b+1)!}\sum_{k=1}^{a-b}\binom{b+k}{k-1}^{-1}\tag{3.2}\\ &=\frac{1}{(b+1)!}\sum_{k=1}^{a-b}(b+k+1)\int_0^1z^{k-1}(1-z)^{b+1}\,dz\tag{3.3}\\ &=\frac{1}{b!}\int_{0}^{1}(1-z)^{b+1}\sum_{k=1}^{a-b}z^{k-1}\,dz\\ &\qquad+\frac{1}{(b+1)!}\int_{0}^{1}(1-z)^{b+1}\sum_{k=1}^{a-b}kz^{k-1}\,dz\\ &=\frac{1}{b!}\int_{0}^{1}(1-z)^{b}\left(1-z^{a-b}\right)\,dz\\ &\qquad+\frac{1}{(b+1)!}\int_{0}^{1}(1-z)^{b+1}\left(\frac{d}{dz}\sum_{k=1}^{a-b}z^k\right)\,dz\tag{3.4}\\ &=\frac{1}{b!}\left(\frac{1}{b+1}-\int_{0}^{1}z^{a-b}(1-z)^b\,dz\right)\\ &\qquad+\frac{1}{(b+1)!}\int_{0}^{1}(1-z)^{b+1}\left(\frac{d}{dz}\left(\frac{z\left(1-z^{a-b}\right)}{1-z}\right)\right)dz\\ &=\frac{1}{b!}\left(\frac{1}{b+1}-\frac{(a-b)!b!}{(a+1)!}\right)+\frac{1}{(b+1)!}\int_{0}^{1}(1-z)^{b+1}\\ &\qquad\qquad\cdot\left(\frac{1}{1-z}-\frac{(a-b+1)z^{a-b}}{1-z}+\frac{z\left(1-z^{a-b}\right)}{(1-z)^2}\right)dz\tag{3.5}\\ &=\frac{1}{(b+1)!}-\frac{(a-b)!}{(a+1)!}+\frac{1}{(b+1)!}\int_{0}^1(1-z)^b\,dz\\ &\qquad-\frac{a-b+1}{(b+1)!}\int_{0}^{1}z^{a-b}(1-z)^b\,dz+\frac{1}{(b+1)!}\int_{0}^{1}z(1-z)^{b-1}\,dz\\ &\qquad-\frac{1}{(b+1)!}\int_{0}^1z^{a-b+1}(1-z)^{b-1}\,dz\\ &=\frac{1}{(b+1)!}-\frac{(a-b)!}{(a+1)!}+\frac{1}{(b+1)!}\,\frac{1}{b+1}\\ &\qquad-\frac{a-b+1}{(b+1)!}\,\frac{(a-b)!b!}{(a+1)!}+\frac{1}{(b+1)!}\,\frac{1}{b(b+1)}\\ &\qquad-\frac{1}{(b+1)!}\,\frac{(a-b+1)!(b-1)!}{(a+1)!}\tag{3.6}\\ &=\frac{1}{(b+1)!}\left(1+\frac{1}{b+1}+\frac{1}{b(b+1)}\right)\\ &\qquad-\frac{(a-b)!}{(a+1)!}\left(1+\frac{a-b+1}{b+1}+\frac{a-b+1}{b(b+1)}\right)\tag{3.7}\\ &=\frac{1}{(b+1)!}\,\frac{b^2+2b+1}{b(b+1)}\\ &\qquad-\frac{(a-b)}{(a+1)!}\,\frac{\left(b^2+b\right)+\left(ab-b^2+b\right)+\left(a-b+1\right)}{b(b+1)}\\ &=\frac{1}{b\,b!}-\frac{(a-b)!}{(a+1)!}\,\frac{(b+1)(a+1)}{b(b+1)}\\ &\,\,\color{blue}{=\frac{1}{b\cdot b!}-\frac{(a-b)!}{b\cdot a!}} \end{align*} and the claim (1) follows.

Comment:

• In (3.1) we change the order of summation $k\to a-b-k+1$.

• In (3.2) we expand numerator and denominator with $(b+1)!$ and write the terms as binomial coefficients.

• In (3.3) we apply the identity (2).

• In (3.4) we apply the finite geometric sum formula in the left-hand integral, cancel the factor $1-z$. In the right-hand integral we write the sum using the differential operator $\frac{d}{dx}$ to conveniently prepare the next finite geometric series expansion which is done in the next step.

• In (3.5) we do the differentiation and split the integrals in the next step.

• In (3.5) and (3.6) we use an identity of the Beta function in the form \begin{align*} \int_{0}^1z^a(1-z)^b &= B(a+1,b+1)=\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}\\ &=\frac{a!b!}{(a+b+1)!} \end{align*} and calculate also the other integrals.

• In (3.7) and the following lines we collect terms and simplify.

We seek to show that

$$\sum_{k=1}^{a-b} \frac{(a-b-k)!}{(a+1-k)!} = \frac{1}{b} \left[\frac{1}{b!}-\frac{(a-b)!}{a!}\right]$$

Recall from MSE 4316307 the following identity which was proved there: with $1\le k\le n$

$${n\choose k}^{-1} = k [z^n] \log\frac{1}{1-z} (-1)^{n-k} (1-z)^{n-k}.$$

We thus have with positive integers $a,b$ where $a-b\ge 1$ that

$$\sum_{k=1}^{a-b} \frac{(a-b-k)!}{(a+1-k)!} = \sum_{k=0}^{a-b-1} \frac{k!}{(b+1+k)!} = \frac{1}{(b+1)!} \sum_{k=0}^{a-b-1} {b+1+k\choose k}^{-1} \\ = \frac{1}{(b+1)!} + \frac{1}{(b+1)!} \sum_{k=1}^{a-b-1} {b+1+k\choose k}^{-1} \\ = \frac{1}{(b+1)!} + \frac{1}{(b+1)!} \sum_{k=1}^{a-b-1} k [z^{b+1+k}] \log\frac{1}{1-z} (-1)^{b+1} (1-z)^{b+1}.$$

We may lower $k$ to zero because there is zero contribution and get for the sum term

$$\sum_{k=0}^{a-b-1} k [z^{b+1+k}] \log\frac{1}{1-z} (-1)^{b+1} (1-z)^{b+1} \\ = \sum_{k=b+1}^a (k-(b+1)) [z^k] \log\frac{1}{1-z} (-1)^{b+1} (1-z)^{b+1}.$$

Two pieces

We thus require two pieces, the first is

$$[w^m] \frac{1}{1-w} \sum_{k\ge 0} w^k k [z^k] \log\frac{1}{1-z} (-1)^{b+1} (1-z)^{b+1}.$$

This is

$$[w^{m-1}] \frac{1}{1-w} \left.\left( \log\frac{1}{1-z} (z-1)^{b+1} \right)'\right|_{z=w} \\ = [w^{m-1}] \frac{1}{1-w} \left.\left(- (z-1)^b + (b+1) \log\frac{1}{1-z} (z-1)^b\right) \right|_{z=w} \\ = [w^{m-1}] \left(((w-1)^{b-1} - (b+1) \log\frac{1}{1-w} (w-1)^{b-1}\right).$$

The second main piece is

$$- (b+1) [w^m] \frac{1}{1-w} \sum_{k\ge 0} w^k [z^k] \log\frac{1}{1-z} (-1)^{b+1} (1-z)^{b+1} \\ = (b+1) [w^m] \log\frac{1}{1-w} (w-1)^b.$$

Evaluating the pieces at $m=a$ and $m=b$

Evaluating at $m=a$ and $m=b$ we get for the first one

$$- \frac{b+1}{a-b} {a-1\choose a-b}^{-1}$$

and the second one

$$1 - (b+1) [w^{b-1}] \log\frac{1}{1-w} (w-1)^{b-1}.$$

Evaluate the second piece again at $m=a$ and $m=b$ we find

$$\frac{b+1}{a-b} {a\choose a-b}^{-1}$$

and

$$(b+1) [w^b] \log\frac{1}{1-w} (w-1)^b.$$

We evidently require

$$(-1)^b \; \underset{w}{\mathrm{res}} \; \frac{1}{w^{b+1}} (1-w)^b \log\frac{1}{1-w}$$

This was also evaluated at the cited link and found to be $-H_b.$

Collecting everything

We obtain at last for the sum component

$$- \frac{b+1}{a-b} {a-1\choose a-b}^{-1} -1 - (b+1) H_{b-1} + \frac{b+1}{a-b} {a\choose a-b}^{-1} + (b+1) H_b \\ = \frac{1}{b} - \frac{b+1}{a-b} {a\choose b}^{-1} \frac{a}{b} + \frac{b+1}{a-b} {a\choose b}^{-1} = \frac{1}{b} + \frac{b+1}{a-b} {a\choose b}^{-1} \frac{b-a}{b}.$$

We get for the complete sum

$$\frac{1}{(b+1)!} + \frac{1}{b\times (b+1)!} - \frac{(a-b)!}{b\times a!},$$

which is the claim.