# A series identity

Let $f_n(x)=\prod_{k=0}^n \frac{1}{x+k}$.

I need to show for every $x \in \mathbb{R}, x>0$:

$\sum\limits_{n=0}^{\infty} f_n(x) = e\sum\limits_{n=0}^{\infty} \frac{(-1)^n}{(x+n)n!}$

What I have noticed: they have the same evaluation for $x=1$: they both go to $e-1$. Also, the left one is easy to evaluate for all natural numbers.

I'm quite sure there is some smart move with Taylor expansion, even though it doesnt't seem.

• have you tried to take logarithms? Feb 4, 2017 at 9:42
• No, but how can you solve it that way? Logarithms well behave with products, not with sums, no? Feb 4, 2017 at 9:46
• i don'T claim this solves the problem, it is just an idea :-D Feb 4, 2017 at 9:49

We have $$S=\sum_{n\geq0}\prod_{k=0}^{n}\frac{1}{x+k}=\sum_{n\geq0}\frac{\Gamma\left(x\right)}{\Gamma\left(x+1+n\right)}$$ $$=\sum_{n\geq0}\frac{1}{n!}B\left(x,n+1\right)=\int_{0}^{1}t^{x-1}\sum_{n\geq0}\frac{\left(1-t\right)^{n}}{n!}dt=e\int_{0}^{1}t^{x-1}e^{-t}dt=\color{red}{e\gamma\left(x,1\right)}$$ where $\gamma\left(x,a\right)$ is the incomplete Gamma function. The claim follows using the representation of $\gamma\left(x,a\right)$ as a confluent hypergeometric function of the first kind. Just for completeness $$S=ex^{-1}\,_{1}F_{1}\left(x;1+x;-1\right)=ex^{-1}\sum_{n\geq0}\frac{\left(x\right)_{n}}{\left(x+1\right)_{n}}\frac{\left(-1\right)^{n}}{n!}$$ $$=ex^{-1}\sum_{n\geq0}\frac{x\left(x+1\right)\cdots\left(x+n-1\right)}{\left(x+1\right)\cdots\left(x+n-1\right)\left(x+n\right)}\frac{\left(-1\right)^{n}}{n!}=e\sum_{n\geq0}\frac{\left(-1\right)^{n}}{\left(x+n\right)n!}.$$

• Nice advanced math, step-by-step. (+1). Feb 4, 2017 at 17:04

It is enough to check that the LHS and the RHS have the same residues at $x=0,-1,-2,-3,\ldots$

Obviously $$\text{Res}\left(e\sum_{n\geq 1}\frac{(-1)^n}{(x+n)n!},x=-m\right)=\frac{(-1)^m}{m!}e\tag{1}$$ while $$\text{Res}\left(\sum_{n\geq 0}\frac{1}{x\cdots(x+n)},x=-m\right)=\frac{(-1)^m}{m!}\sum_{k\geq 0}\frac{1}{k!}\tag{2}$$ that is exacty the same.

• Can you link the theorem you used to conclude? Feb 4, 2017 at 14:43
• @ZenoCozeno: sure. math.stackexchange.com/questions/581162/… Feb 4, 2017 at 15:13
• the functions can still differ by a constant right (which is zero in this case)? Feb 4, 2017 at 15:23
• @tired: of course, but the difference is zero by performing an explicit evaluation at $x=1$. Feb 4, 2017 at 15:42

Disclaimer: This was not an answer.

As a possible hint, observe that $$f(x)=\sum_{n=0}^\infty f_n(x)=\sum_{n=0}^\infty\frac 1{x(x+1)\cdots(x+n)}$$ and $$f(x)-f(x+1)=\sum_{n=1}^\infty\frac n{x(x+1)\cdots(x+n)}=\sum_{n=1}^\infty n f_n(x)$$ On the other hand, we know that $$f_n(x+1)=\prod_{k=0}^n \frac{1}{x+k+1}=xf_{n+1}(x)$$ Consequently $$f(x+1)=\sum_{n=0}^\infty f_n(x+1)=x\sum_{n=1}^\infty f_n(x)=x f(x)-1$$ Thus $$1+(1-x)\sum_{n=0}^\infty f_n(x)=\sum_{n=0}^\infty n f_n(x)$$

Let $g(t)=(1+t)^{-x}$ then $$\mathcal {D_t}^n g(0)=(-1)^nx(x+1)\cdots(x+n)$$ I thought these relationships could be useful.