Prove $\sum_n^{\infty} \prod_{k=0}^n \dfrac{1}{x+k} = e \sum_ n^{\infty} \dfrac{(-1)^n}{(x+n)n!}$ Let $$f_n(x) = \prod_{k=0}^n \dfrac{1}{x+k}.$$ Show that $$\sum_{n=0}^{\infty} f_n(x) = e \sum_ {n=0}^{\infty} \dfrac{(-1)^n}{(x+n)n!}.$$
 A: We have $f_n(x)=\sum_{m=0}^n c_m/(x+m)$ for some $c_m$'s. To compute them, multiply by $x+m$, set $x=-m$, and get $c_m=(-1)^m\frac{1}{m!(n-m)!}$. We thus have
$$\sum_{n=0}^Nf_n=\sum_{N\leq n\leq m\leq0}\frac{(-1)^m}{m!(n-m)!(x+m)}=$$
(setting $k=n-m$)
$$=\sum_{N\leq m\leq0,\,N-m\leq k\leq0}\frac{1}{k!}\frac{(-1)^m}{m!(x+m)}.$$
Now take the limit $N\to\infty$.
A: Using the partial fraction identity that was proved by a straightforward induction technique at this MSE post, we have that
$$\prod_{k=0}^n \frac{1}{x+k} =
\frac{1}{n!} \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{x+k}.$$
Now to compute $$\sum_{n=0}^\infty \prod_{k=0}^n \frac{1}{x+k}$$ we ask about the coefficient of $$\frac{1}{x+k}$$ taking into account all terms of the series. We see immediately that all products with $n\ge k$ include this term, so the coefficient is
$$(-1)^k \sum_{n\ge k} \frac{1}{n!} \binom{n}{k}
= \frac{(-1)^k}{k!} \sum_{n\ge k} \frac{1}{(n-k)!}  =
e\frac{(-1)^k}{k!}.$$
Now summing for $k\ge 0$ we get the desired result
$$e\sum_{k=0}^\infty \frac{(-1)^k}{k!} \frac{1}{x+k}.$$
Interestingly enough the term
$$g_n(x) = \prod_{k=1}^n \frac{1}{x+k}$$ can also be evaluated using Mellin transforms.
(Drop the factor $1/x$ for the moment to keep the Mellin integral simple -- no pole at zero.)
We get $$g_n^*(s) = \mathfrak{M}(g_n(x); s)
= \int_0^\infty \prod_{k=1}^n \frac{1}{x+k} x^{s-1} dx$$
which gives (use a keyhole contour with the slot on the real axis, which is also the branch cut of the logarithm for $x^{s-1}$) $$g_n^*(s) (1-e^{2\pi i(s-1)}) = 2\pi i
\sum_{q=1}^n \operatorname{Res}(g_n(x); x=-q).$$
The sum of the residues is
$$ \sum_{q=1}^n \operatorname{Res}(g_n(x); x=-q) \\=
\sum_{q=1}^n (-q)^{s-1}
 \prod_{k=1}^{q-1} \frac{1}{-q+k} \prod_{k=q+1}^n \frac{1}{-q+k}
= \sum_{q=1}^n e^{i\pi(s-1)} q^{s-1} \frac{(-1)^{q-1}}{(q-1)!} \frac{1}{(n-q)!} \\=
- e^{i\pi s} \sum_{q=1}^n q^s \frac{(-1)^{q-1}}{q!} \frac{1}{(n-q)!}
= -\frac{e^{i\pi s}}{n!}\sum_{q=1}^n q^s (-1)^{q-1} \binom{n}{q}.$$
This gives
$$g_n^*(s) = - \frac{1}{n!} 
\frac{2\pi i \times e^{i\pi s}}{1-e^{2\pi i(s-1)}} \sum_{q=1}^n q^s (-1)^{q-1} \binom{n}{q}
\\ = - \frac{1}{n!} 
\frac{2\pi i }{e^{-\pi i s} - e^{\pi i s}} \sum_{q=1}^n q^s (-1)^{q-1} \binom{n}{q}
= \frac{1}{n!} 
\frac{\pi}{\sin(\pi s)} \sum_{q=1}^n q^s (-1)^{q-1} \binom{n}{q}.$$
We apply Mellin inversion to recover $g_n(x)$ with the Mellin inversion integral being
$$\frac{1}{2\pi i}\int_{1/2-i\infty}^{1/2+i\infty} g_n^*(s)/x^s ds,$$
shifting to the left to recover an expansion about zero.
The residue at $s=0$ is special, it has the value
$$\frac{1}{n!} \sum_{q=1}^n  (-1)^{q-1} \binom{n}{q} = \frac{1}{n!} .$$
The remaining residues at the negative integers $-p$ contribute
$$\frac{1}{n!} \sum_{p=1}^\infty (-1)^p x^p
\sum_{q=1}^n \frac{1}{q^p} (-1)^{q-1} \binom{n}{q}
= \frac{1}{n!} \sum_{q=1}^n (-1)^{q-1} \binom{n}{q}
\sum_{p=1}^\infty  \frac{(-1)^p\times x^p}{q^p} \\=
\frac{1}{n!} \sum_{q=1}^n (-1)^{q-1} \binom{n}{q}
\frac{-x/q}{1+x/q}
= \frac{1}{n!} \sum_{q=1}^n (-1)^q \binom{n}{q} \frac{x}{x+q}.$$
Including the residue at zero we thus obtain
$$ g_n(x) = \frac{1}{n!} + \frac{1}{n!} \sum_{q=1}^n (-1)^q \binom{n}{q} \frac{x}{x+q}
= \frac{1}{n!} \sum_{q=0}^n (-1)^q \binom{n}{q} \frac{x}{x+q}.$$
Since $f_n(x) = 1/x \times g_n(x)$ we get that
$$f_n(x) =  \frac{1}{n!} \sum_{q=0}^n (-1)^q \binom{n}{q} \frac{1}{x+q}.$$
Observation Aug 25 2014. The Mellin transform calculation is little more than a reworked computation of the partial fraction decomposition of the product term by residues and is in fact not strictly necessary here. An example of this very simple technique (no transforms) is at this MSE link.
