Sum of reciprocals of product of consecutive integers

Note that, since $$ \bbox[lightyellow] { x^{\;\overline {\,a + b\,} } = x^{\;\overline {\,a\,} } \left( {x + a} \right)^{\;\overline {\,b\,} } = x^{\;\overline {\,a\,} } \left( {x + a + b - 1} \right)^{\,\underline {\,b\,} } \quad {\rm and}\quad x^{\;\overline {\,0\,} } = 1 }$$

then for negative exponents the Rising and Falling Factorial are defined as: $$ \bbox[lightyellow] { {1 \over {x^{\;\overline {\,m\,} } }} = \left( {x - 1} \right)^{\,\underline {\, - m\,} } \quad {1 \over {x^{\,\underline {\,m\,} } }} = \left( {x + 1} \right)^{\;\overline {\, - m\,} } }$$

This definition is quite "natural", since it preserves most of the properties of the Rising and Falling factorials as defined for positive exponents, and which includes the "exponent split" introduced above, the definition through the Gamma function, etc.
It is the "standard" definition that you find e.g. in the Wikipedia article cited above, and in Wolfram site.
A thoroughful, unsurpassed, explanation of Rising/Falling factorials, as well as of the Difference, definite and indefinite sum, is provided in the renowned Concrete Mathematics.

Consider now the Forward Finite Difference (Delta) defined as: $$ \bbox[lightyellow] { \Delta _{\,x} F(x) = F(x + 1) - F(x)\quad \Delta _{\,x} ^m F(x) = \Delta _{\,x} \left( {\Delta _{\,x} ^{m - 1} F(x)\;} \right) }$$ and its functional inverse: the Anti-Difference or Indefinite Sum $$ \bbox[lightyellow] { \eqalign{ & \Delta _{\,x} F(x) = f(x)\quad \Leftrightarrow \quad \Delta _{\,x} ^{ - 1} f(x) = \quad \sum\nolimits_x {f(x)} = F(x) + c\quad \Leftrightarrow \cr & \Leftrightarrow \quad \sum\nolimits_{\;x\, = \,a}^{\;b} {f(x)} = \sum\limits_{a\, \le \,x\, \le \,b - 1} {f(x)} = F(b) - F(a) \cr} }$$ where the last identity is just denoting a telescoping sum, once you replace $f(x)$ with $F(x+1)-F(x)$.
It is evident the similarity with the derivative vs. the indefinite and definite integral .
Like for the indefinite integral, in fact, $c$ represents a constant or, more generally, any periodic function of $x$ with period (one of the periods) =1.

Now, the formulas for the finite differences of the Falling factorial are easy to be deducted, as to be $$ \bbox[lightyellow] { \eqalign{ & \Delta _{\,x} ^m \;x^{\,\underline {\,n\,} } = \nabla _{\,x} ^m \;\left( {x + m} \right)^{\,\underline {\,n\,} } = n^{\,\underline {\,m\,} } x^{\,\underline {\,n - m\,} } \cr & \Delta _{\,x} ^m \;x^{\,\overline {\,n\,} } = \nabla _{\,x} ^m \;\left( {x + m} \right)^{\,\overline {\,n\,} } = n^{\,\underline {\,m\,} } \left( {x + m} \right)^{\,\overline {\,n - m\,} } = n^{\,\underline {\,m\,} } \left( {x + n - 1} \right)^{\,\underline {\,n - m\,} } \cr} } \tag{1}$$ and it is also easy to check that they hold either for positive and for negative values of $n$, given the definition above.

But they hold as well for negative values of $m$, and for $m=-1$ in particular, we can check that, for the Falling Factorial we have $$ \bbox[lightyellow] { \sum\nolimits_{\;x} {x^{\,\underline {\,n\,} } } = \left\{ {\matrix{ {{1 \over {n + 1}}\;x^{\,\underline {\,n + 1\,} } + c} & { - 1 \ne n} \cr {\psi (x + 1) + c} & { - 1 = n} \cr } } \right.\quad \leftrightarrow \quad \int {x^{\,n} dx} = \left\{ {\matrix{ {{1 \over {n + 1}}\;x^{\,n + 1} + c} & { - 1 \ne n} \cr {\ln (x) + c} & { - 1 = n} \cr } } \right. } \tag{2.a}$$ while for the Rising Factorial it is $$ \bbox[lightyellow] { \sum\nolimits_{\;x} {x^{\,\overline {\,n\,} } } = \sum\nolimits_{\;x} {\left( {x + n - 1} \right)^{\,\underline {\,n\,} } } = \left\{ {\matrix{ {{1 \over {n + 1}}\;\left( {x - 1} \right)^{\,\overline {\,n + 1\,} } + c} & { - 1 \ne n} \cr {\psi (x - 1) + c} & { - 1 = n} \cr } } \right. } \tag{2.b}$$ where $\psi(x)$ is the Digamma function (which, interestingly, completes the analogy by pairing with the log).

Example

$$ \bbox[lightyellow] { x^{\,\underline {\, - 2\,} } = {1 \over {\left( {x + 1} \right)^{\,\overline {\,2\,} } }} = {1 \over {\left( {x + 1} \right)\left( {x + 2} \right)}} }$$ then, for the difference $$ \bbox[lightyellow] { \eqalign{ & \Delta _{\,x} x^{\,\underline {\, - 2\,} } = {1 \over {\left( {x + 2} \right)\left( {x + 3} \right)}} - {1 \over {\left( {x + 1} \right)\left( {x + 2} \right)}} = \cr & = - {2 \over {\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)}} = - 2{1 \over {\left( {x + 1} \right)^{\,\overline {\,3\,} } }} = - 2x^{\,\underline {\, - 3\,} } \cr} }$$ but it is also $$ \bbox[lightyellow] { - 2x^{\,\underline {\, - 3\,} } = \Delta _{\,x} \left( {x^{\,\underline {\, - 2\,} } + 1} \right) = \Delta _{\,x} \left( {x^{\,\underline {\, - 2\,} } + \sin \left( {2\pi x} \right)} \right) = \Delta _{\,x} \left( {x^{\,\underline {\, - 2\,} } + x\bmod 1} \right) = \cdots }$$ and for the sum $$ \bbox[lightyellow] { \eqalign{ & \sum\limits_{0\, \le \,x\, \le \,n - 1} {x^{\,\underline {\, - 3\,} } } = - {1 \over 2}\sum\limits_{0\, \le \,x\, \le \,n - 1} {\left( {{1 \over {\left( {x + 2} \right)\left( {x + 3} \right)}} - {1 \over {\left( {x + 1} \right)\left( {x + 2} \right)}}} \right)} = \cr & = - {1 \over 2}\left( {{1 \over {\left( {n + 1} \right)\left( {n + 2} \right)}} - {1 \over {\left( {0 + 1} \right)\left( {0 + 2} \right)}}} \right) = \cr & = - {1 \over 2}\left( {n^{\,\underline {\, - 2\,} } - 0^{\,\underline {\, - 2\,} } } \right) \cr} }$$

In addition, the example provided in Thorgott's answer, translates to: $$ \bbox[lightyellow] { \eqalign{ & \sum\limits_{1\, \le \,k\,} {{1 \over {\prod\limits_{0\, \le \,x\, \le \,n} {\left( {k + i} \right)} }}} = \sum\limits_{1\, \le \,k\,} {{1 \over {k^{\,\overline {\,n + 1\,} } }}} = \sum\limits_{0\, \le \,k\,} {{1 \over {\left( {k + 1} \right)^{\,\overline {\,n + 1\,} } }}} = \cr & = \sum\limits_{0\, \le \,k\,} {k^{\,\underline {\, - n - 1\,} } } = - {1 \over n}\left[ {k^{\,\underline {\, - n\,} } } \right]_{k = 0}^{\;\infty } = - {1 \over n}\left[ {{1 \over {\left( {k + 1} \right)^{\,\overline {\,n\,} } }}} \right]_{k = 0}^{\;\infty } = {1 \over n}{1 \over {1^{\,\overline {\,n\,} } }} = {1 \over n}{1 \over {n!}} \cr} }$$

Note that

$$\frac 1{r^\overline{m}}=\frac 1{m-1}\left[\frac 1{r^\overline{m-1}}-\frac 1{(r+1)^\overline{m-1}}\right]$$

Summing to $n$ gives $$\begin{align} \sum_{r=1}^n \frac 1{r^\overline{m}} &=\frac 1{m-1}\left[\frac 1{1^\overline{m-1}}-\frac 1{(n+1)^\overline{m-1}}\right]\\ &=\frac 1{m-1}\left[\frac 1{(m-1)!}-\frac 1{(n+1)^\overline{m-1}}\right]\tag{3}\end{align}$$

and as $n\to \infty$, we have

$$\sum_{r=1}^\infty \frac 1{r^\overline{m}}=\frac 1{(m-1)(m-1)!}=\frac 1{m!-(m-1)!}$$

Hence the only questions remaining are whether $(2)$ can be expressed as an extension of $(1)$, and also, whether there is a neat form for $(3)$.

If we denote $\dfrac 1{r^\overline{m}}$ as $r^{-\overline{m}}$, then $(3)$ can be written as $$\sum_{r=1}^n r^{-\overline{m}}=\frac {1^{-\overline{m-1}}-(n+1)^{-\overline{m-1}}}{m-1}\tag{4}$$ which is analogous to $$\int_1^{n+1} x^{-m}dx=\left[\frac {x^{-m+1}}{-m+1}\right]_1^{n+1}=\frac {1^{-(m-1)}-(n+1)^{-(m-1)}}{m-1}\tag{4A}$$ It is interesting to note that for both $(1),(3)$, the summation is taken from $1$ to $n$, but for $(1$A$)$, the integration is taken from $0$ to $n$, and for $(4$A$)$ from $1$ to $n+1$.

Why should this be so?

NB - this was previously in the addendum but is now posted as a solution for clarity as it answers the original question.

Addendum

See this for a further generalisation.

If we define the compact notation of the reciprocal of a modified rising factorial as $$r^{-\stackrel{\langle d\rangle}{\overline{m}}}=\frac 1{\underbrace{r(r+d)(r+2d)\cdots ((m-1)d)}_{m\text{ terms}}}$$ then the sum for $r=1$ to $n$ can be written as

$$\sum_{r=1}^n r^{-\stackrel{\langle d\rangle}{\overline{m}}} = \frac {1^{-\stackrel{\langle d\rangle}{\overline{(m-1)}}}-(n+1)^{-\stackrel{\langle d\rangle}{\overline{(m-1)}}}}{(m-1)d}$$

I will prove the identity $$F(n)=\sum\limits_{k=1}^{\infty}\frac{1}{\prod\limits_{i=0}^{n}(k+i)}=\frac{1}{n!n}$$ Proof: To prove this, rewrite the sum using the identity $$\frac{1}{\prod\limits_{i=0}^n(k+i)}=\frac{1}{n!}\sum\limits_{i=0}^n{{n}\choose{i}}\frac{(-1)^i}{k+i}$$ This identity might be derived using partial fraction decomposition and taking limits or simply be confirmed by induction (see here: https://mathoverflow.net/questions/193611/binomial-coefficient-identity/193649#193649). Then \begin{align}F(n)&=\sum\limits_{k=1}^{\infty}\frac{1}{\prod\limits_{i=0}^{n}(k+i)}=\sum\limits_{k=1}^{\infty}\Bigg[\frac{1}{n!}\sum\limits_{i=0}^n{{n}\choose{i}}\frac{(-1)^i}{k+i}\Bigg]\\ &=\frac{1}{n!}\sum\limits_{k=1}^{\infty}\sum\limits_{i=0}^{n-1}(-1)^i{{n-1}\choose{i}}\Bigg(\frac{1}{k+i}-\frac{1}{k+i+1}\Bigg)\end{align} The latter equality follows by comparing the coefficients of $\frac{1}{k+i}$ in the resulting sum (mostly $-(-1)^{i-1}{{n-1}\choose{i-1}}+(-1)^i{{n-1}\choose{i}}=(-1)^i{{n}\choose{i}}$ due to Pascal's Identity). Now \begin{align} F(n)&=\frac{1}{n!}\sum\limits_{k=1}^{\infty}\sum\limits_{i=0}^{n-1}(-1)^i{{n-1}\choose{i}}\Bigg(\frac{1}{k+i}-\frac{1}{k+i+1}\Bigg)\\ &=\frac{1}{n!}\lim\limits_{r\rightarrow\infty}\sum\limits_{k=1}^{r}\sum\limits_{i=0}^{n-1}(-1)^i{{n-1}\choose{i}}\Bigg(\frac{1}{k+i}-\frac{1}{k+i+1}\Bigg)\\ &=\frac{1}{n!}\sum\limits_{i=0}^{n-1}(-1)^i{{n-1}\choose{i}}\lim\limits_{r\rightarrow\infty}\sum\limits_{k=1}^{r}\Bigg(\frac{1}{k+i}-\frac{1}{k+i+1}\Bigg)\\ &=\frac{1}{n!}\sum\limits_{i=0}^{n-1}(-1)^i{{n-1}\choose{i}}\lim\limits_{r\rightarrow\infty}\Bigg(\frac{1}{1+i}-\frac{1}{r+i+1}\Bigg)\\ &=\frac{1}{n!}\sum\limits_{i=0}^{n-1}(-1)^i{{n-1}\choose{i}}\frac{1}{1+i} \end{align} Utilize $(1+i){{n}\choose{1+i}}=n{{n-1}\choose{i}}$ to rewrite this \begin{align} F(n)&=\frac{1}{n!}\sum\limits_{i=0}^{n-1}(-1)^i{{n-1}\choose{i}}\frac{1}{1+i}\\ &=\frac{1}{n!n}\sum\limits_{i=0}^{n-1}(-1)^i{{n}\choose{i+1}}\\ &=\frac{1}{n!n}\sum\limits_{j=1}^{n}(-1)^{j-1}{{n}\choose{j}}\\ &=-\frac{1}{n!n}\sum\limits_{j=1}^{n}(-1)^j{{n}\choose{j}}\\ &=-\frac{1}{n!n}\Bigg[\sum\limits_{j=0}^{n}(-1)^j{{n}\choose{j}}-(-1)^0{{n}\choose{0}}\Bigg]\\ &=-\frac{1}{n!n}(-1)\\ &=\frac{1}{n!n} \end{align} As seen in the proof, the following identity holds for partial sums: $$\sum\limits_{k=1}^{r}\frac{1}{\prod\limits_{i=0}^{n}(k+i)}=\frac{1}{n!}\sum\limits_{i=0}^{n-1}(-1)^i{{n-1}\choose{i}}\Bigg(\frac{1}{1+i}-\frac{1}{r+i+1}\Bigg)$$ However, I am not sure how this sum can further be simplified.