2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ have you tried to take logarithms? $\endgroup$
    – tired
    Feb 4, 2017 at 9:42
  • $\begingroup$ No, but how can you solve it that way? Logarithms well behave with products, not with sums, no? $\endgroup$
    – ZenoCozeno
    Feb 4, 2017 at 9:46
  • $\begingroup$ i don'T claim this solves the problem, it is just an idea :-D $\endgroup$
    – tired
    Feb 4, 2017 at 9:49

3 Answers 3

3
$\begingroup$

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 $$S=ex^{-1}\,_{1}F_{1}\left(x;1+x;-1\right)=ex^{-1}\sum_{n\geq0}\frac{\left(x\right)_{n}}{\left(x+1\right)_{n}}\frac{\left(-1\right)^{n}}{n!}$$ $$=ex^{-1}\sum_{n\geq0}\frac{x\left(x+1\right)\cdots\left(x+n-1\right)}{\left(x+1\right)\cdots\left(x+n-1\right)\left(x+n\right)}\frac{\left(-1\right)^{n}}{n!}=e\sum_{n\geq0}\frac{\left(-1\right)^{n}}{\left(x+n\right)n!}.$$

$\endgroup$
1
  • 1
    $\begingroup$ Nice advanced math, step-by-step. (+1). $\endgroup$ Feb 4, 2017 at 17:04
2
$\begingroup$

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.

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.