TITLE: Sum of reciprocals of product of consecutive integers
QUESTION [6 upvotes]: We know, for instance, that
$$\sum_{r=1}^n r^\overline{3}=\sum_{r=1}^nr(r+1)(r+2)=\frac {n(n+1)(n+2)(n+3)}4=\frac {n^\overline{4}}4$$
and, in general, 
$$\sum_{r=1}^n r^\overline{m}=\frac {n^\overline{m+1}}{m+1}\tag{1}$$
which is analogous to 
$$\int_0^n x^m dx=\frac {n^{m+1}}{m+1}\tag{1A}$$

but is there is correspondingly neat form (and analogy) for sum of reciprocals, e.g.
  $\displaystyle\sum_{r=1}^n \frac 1{r(r+1)(r+2)}$
  and in general
$$\sum_{r=1}^n \frac 1{r^\overline{m}}\tag{2}$$
  ?
Or, alternatively, can it be shown that $(2)$ is an extension of $(1)$?  
Also, can it be shown easily that 
  $$\sum_{r=1}^\infty\frac 1{r^\overline{m}}=\frac 1{(m-1)(m-1)!}=\frac 1{m!-(m-1)!}$$
  ?


NB: Here the notation for the rising factorial is used, i.e. $x^\overline{m}=\overbrace{x(x+1)(x+2)\cdots (x+m-1)}^{m\text{ terms}}$.

Addendum
See this for a further generalisation.

REPLY [2 votes]: 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:
$$ \bbox[lightyellow] {  
\eqalign{
  & \sum\limits_{1\, \le \,k\,} {{1 \over {\prod\limits_{0\, \le \,x\, \le \,n} {\left( {k + i} \right)} }}}  = \sum\limits_{1\, \le \,k\,} {{1 \over {k^{\,\overline {\,n + 1\,} } }}}  = \sum\limits_{0\, \le \,k\,} {{1 \over {\left( {k + 1} \right)^{\,\overline {\,n + 1\,} } }}}  =   \cr 
  &  = \sum\limits_{0\, \le \,k\,} {k^{\,\underline {\, - n - 1\,} } }  =  - {1 \over n}\left[ {k^{\,\underline {\, - n\,} } } \right]_{k = 0}^{\;\infty }  =  - {1 \over n}\left[ {{1 \over {\left( {k + 1} \right)^{\,\overline {\,n\,} } }}} \right]_{k = 0}^{\;\infty }  = {1 \over n}{1 \over {1^{\,\overline {\,n\,} } }} = {1 \over n}{1 \over {n!}} \cr} 
 }$$