How do I evaluate this limit: $\lim_{n\to+\infty}\sum_{k=1}^{n} \frac{1}{k(k+1)\cdots(k+m)}$? How do I evaluate this limit?
$$\lim_{n\to+\infty}\sum_{k=1}^{n} \frac{1}{k(k+1)\cdots(k+m)} \qquad (m=1,2,3,\cdots)$$
Thanks in advance.

Here is an answer without beta functions. Note that the residue of
$$f(z) = \frac{1}{z(z+1)\cdots(z+m)}$$
at $z=-q$ is
$$\mathrm{Res}(f(z); z=-q)
=\frac{1}{(-q)(-q+1)\cdots(-q+(q-1))(-q+(q+1))\cdots(-q+m)}$$
which is
$$\frac{(-1)^q}{q!\times (m-q)!} = \frac{(-1)^q}{m!} {m\choose q}.$$ 
Hence by the residue method for partial fractions we get that
$$f(z) = \frac{1}{m!} \sum_{q=0}^m {m\choose q} (-1)^q \frac{1}{z+q}.$$ 
Apply this to the sum to obtain
$$\sum_{k=1}^\infty \frac{1}{k(k+1)\cdots(k+m)} =
\frac{1}{m!} \sum_{k=1}^\infty \sum_{q=0}^m {m\choose q} (-1)^q \frac{1}{k+q}.$$
Now ask what the coefficient is of the term $1/(k+q) = 1/p$ to get
$$\frac{1}{m!} \sum_{p=1}^\infty \frac{1}{p}\sum_{q=0}^{p-1} {m\choose q} (-1)^q.$$ 
Observe that when $p-1\ge m$ the inner sum becomes zero, so that we are left with
$$\frac{1}{m!} \sum_{p=1}^m \frac{1}{p}\sum_{q=0}^{p-1}  {m\choose q} (-1)^q.$$ 
It therefore remains to evaluate the sum term.
We seek to show that
$$\frac{1}{m} =\sum_{p=1}^m \frac{1}{p}\sum_{q=0}^{p-1}  {m\choose q} (-1)^q.$$
This is the moment to apply what H. Wilf terms the 'snake oil' method in his famous text on generating functions. (Consult this Mathworld Entry.) Introduce the univariate generating function 
$$g(z) = 
\sum_{m\ge 1} z^m \sum_{p=1}^m \frac{1}{p}\sum_{q=0}^{p-1}  {m\choose q} (-1)^q
= \sum_{p=1}^\infty \frac{1}{p}
\sum_{m=p}^\infty z^m \sum_{q=0}^{p-1}  {m\choose q} (-1)^q
\\ = \sum_{p=1}^\infty \frac{z^p}{p}
\sum_{m=0}^\infty z^m \sum_{q=0}^{p-1}  {m+p\choose q} (-1)^q.$$
Now observe that
$$  \sum_{q=0}^{p-1}  {m+p\choose q} (-1)^q = (-1)^{p-1} {m+p-1\choose p-1}$$
as can be seen by induction on $m$, which we now carry out. 
For $m=0$ we get equality with 
$$\sum_{q=0}^{p-1}  {p\choose q} (-1)^q = -(-1)^p = (-1)^{p-1} {p-1\choose p-1}.$$
and for the induction step we have
$$  \sum_{q=0}^{p-1}  {m+1+p\choose q} (-1)^q 
= 1 + \sum_{q=1}^{p-1}  {m+1+p\choose q} (-1)^q
\\= 1 + \sum_{q=1}^{p-1} {m+p\choose q} (-1)^q 
+ \sum_{q=1}^{p-1} {m+p\choose q-1} (-1)^q 
\\= (-1)^{p-1} {m+p-1\choose p-1} 
- \sum_{q=0}^{p-2} {m+p\choose q} (-1)^q 
\\ = (-1)^{p-1} {m+p-1\choose p-1} 
- \sum_{q=0}^{p-1} {m+p\choose q} (-1)^q + (-1)^{p-1} {m+1+p-1\choose p-1}
\\ =  (-1)^{p-1} {m+1+p-1\choose p-1}.$$
Returning to $g(z)$ we thus obtain
$$g(z) =  \sum_{p=1}^\infty \frac{z^p}{p}
\sum_{m=0}^\infty z^m (-1)^{p-1} {m+p-1\choose p-1}
= - \sum_{p=1}^\infty \frac{z^p}{p} (-1)^p \frac{1}{(1-z)^p}$$
by the Newton binomial.
Finally recall that $$\log\frac{1}{1-z} = \sum_{q\ge 1} \frac{z^q}{q}$$
so that $g(z)$ simplifies to
$$ -\log\frac{1}{1+z/(1-z)} = -\log\frac{1-z}{1-z+z} = -\log(1-z)
= \log\frac{1}{1-z}.$$
Therefore $$[z^m] g(z) = \frac{1}{m}$$ and we are done.
There is a similar but not quite identical computation at this MSE link.
