# Asymptotics for a series of products

I am trying to solve the following problem:

Define the following functions for $x>0$: $$f_n(x):=\prod_{k=0}^{n}\frac{1}{x+k}$$

1. Show that the function $$f(x):=\sum_{n=0}^{+\infty}f_n(x)$$ is well defined for $x>0$. Calculate its value in $1$.

2. Study the function $f(x)$ and give asymptotic estimates for $x \to 0^+$ and $x\to +\infty$.

3. Prove that the following equivalence holds: $$f(x)=e \sum_{n=0}^{+\infty}\frac{(-1)^n}{(x+n)n!}$$

I am having a hard time proving the equality in the third point. What I have done for now:

$\textbf{Part 1}$

Using the ratio test, $$\lim_{n\to +\infty}\frac{\prod_{k=0}^{n+1}\frac{1}{x+k}}{\prod_{k=0}^{n}\frac{1}{x+k}}=\lim_{n\to +\infty}\frac{1}{x+n+1}=0$$ the series converges for $x>0$. The value of the function in $1$ is

$$f(1)=\sum_{n=0}^{+\infty}\prod_{k=0}^{n}\frac{1}{k+1}=\sum_{n=0}^{+\infty}\frac{1}{(n+1)!}=e-1$$

$\textbf{Part 2}$

First of all, $f$ is positive for every $x>0$. Its monotonicity is immediate: if $x_2>x_1$,

\begin{align} \quad \qquad \frac{1}{x_2+k}<\frac{1}{x_1+k} \end{align} \\ \implies f(x_2)=\sum_{n=0}^{+\infty}\prod_{k=0}^{n}\frac{1}{x_2+k}\leq\sum_{n=0}^{+\infty}\prod_{k=0}^{n}\frac{1}{x_1+k}=f(x_1)

The general term of the series $f$ must be zero, because it converges; hence in an interval $[M,+\infty)$ with $M>0$

$$||f_n(x) ||_{\infty}=\prod_{k=0}^{n}\frac{1}{M+k}$$ $$\implies \sum_{n=0}^{+\infty}||f_n(x)|| \text{ is convergent}$$

so the series is uniformly convergent on every interval of the type $[M,+\infty)$.

$f$ is asymptotic to $\frac 1x$ for $x\to +\infty$: in fact

$$\lim_{x\to \infty}\frac{f(x)}{\frac{1}{x}}= \lim_{x\to \infty} x\left (\frac{1}{x}+ \sum_{n=1}^{+\infty}\prod_{k=0}^{n}\frac{1}{x+k}\right )= 1$$

because the series converges in a neighbourhood of $+\infty$.

In a neighbourhood of $0$, the function acts similarly: we can notice that

$$\lim_{x\to 0^+}\frac{f(x)}{\frac{1}{x}}=\lim_{x\to 0^+} x\sum_{n=0}^{+\infty}\prod_{k=0}^{n}\frac{1}{x+k}=\lim_{x \to 0^+} x\left (\frac{1}{x}+ \sum_{n=1}^{+\infty}\prod_{k=0}^{n}\frac{1}{x+k}\right )= \lim_{x\to 0^+} 1 + \sum_{n=1}^{+\infty}\prod_{k=1}^{n}\frac{1}{x+k}$$

but $\sum_{n=1}^{+\infty}\prod_{k=1}^{n}\frac{1}{x+k}$ converges in $x=0$ and is continuous, so the limit is

$$\lim_{x\to 0^+}\frac{f(x)}{\frac{1}{x}} = 1+ \sum_{n=1}^{+\infty}\prod_{k=1}^{n}\frac{1}{k}=e$$

hence $f \sim \frac{e}{x}$

Monotonicity and limits of this function imply that $f$ is a bijection of $(0,+\infty)$ in itself.

$\textbf{Part 3}$

I have tried to manipulate the sums: writing a single fraction instead of the product does not seem to work: it leads to

$$\sum_{n=0}^{+\infty}\prod_{k=0}^{n}\frac{1}{x+k}=\frac{1}{x}+\frac{1}{x}\frac{1}{x(x+1)}+\dots=\lim_{n\to +\infty}\frac{\sum_{h=0}^{n}\prod_{k=0}^h(x+k)}{\prod_{k=0}^{n}(x+k)}$$

It does not seem very familiar, even dividing it by $e=\sum_{n=0}^{+\infty}\frac{1}{n!}=f(1)$ Another idea that came to mind was to use the Cauchy product series and the Cauchy series product on the RHS: it leads to

$$\sum_{i=0}^{+\infty}\frac{1}{i!}\sum_{j=0}^{+\infty}\frac{(-1)^j}{(x+j)j!}=\sum_{k=0}^{+\infty}\sum_{l=0}^{k}\frac{(-1)^{k-l}}{(x+k-l)l!(k-l)!}$$

Things seem as complicated as before. Integrating or derivating $f(x)$ term by term would require to know a general form for the integral/derivative of $f_n(x)=\prod_{k=0}^{n}\frac{1}{x+k}$: it does not appear impossible to find it, but I think it would not be of great practical use; moreover, the series does not converge uniformly on the whole interval $(0,+\infty)$. The same goes for the series on the RHS. Working backwards, I thought of finding its integral/series on the interval $[M,+\infty)$ : I obtained

$$\int \left (e\sum_{n=0}^{+\infty}\frac{(-1)^n}{(x+n)n!} \right ) dx =e\sum_{n=0}^{+\infty} \int \frac{(-1)^n}{(x+n)n!} dx=e\sum_{n=0}^{+\infty} \frac{(-1)^n}{n!}\log(x+n)+C$$

I can't get far from here, and I am not even sure if what I have done is correct.

Question: Are the two first parts correct? What could be a good way of proving the equality in the third part?

• It looks like you have a small typo in your demonstration of monotonicity for part $2$. The terms $f(x_2)$ and $f(x_1)$ seem to be in each other's places. Also I believe you could replace $\leq$ with $<$, though that's not important. – Colm Bhandal Dec 27 '15 at 13:54
• @ColmBhandal Thank you for your observation. Got it fixed! – Lonidard Dec 27 '15 at 14:06
• I think your derivation of $f(1)$ is off by $1$, you sum over $1/(n+1)!$, thus resulting in $f(1)=e-1$. – mlk Dec 28 '15 at 8:49
• @mlk True. I'll fix it. – Lonidard Dec 28 '15 at 14:34

For the remaining third point, I would use the formula $$f_n(x)=\frac1{n!}\int_0^1 t^{x-1}(1-t)^{n}dt.$$ Exchanging the order of summation and integration, we get \begin{align*} f(x)&=\int_0^1 t^{x-1} \left(\sum_{n=0}^{\infty}\frac{(1-t)^n}{n!}\right)dt =\int_0^1 t^{x-1}e^{1-t}dt=\\&=e\sum_{k=0}^{\infty}\int_0^1\frac{(-1)^k t^{x-1+k}}{k!}dt=e\sum_{k=0}^{\infty}\frac{(-1)^k}{(x+k)k!}. \end{align*}
• It's obvious once one knows that $\beta(x,n+1)=n!{\prod_{k=0}^{n}\frac{1}{x+k}}$. I did not think about it. Good answer, thank you very much! – Lonidard Dec 20 '15 at 15:16
• Just one doubt: the series does not appear t converge uniformly on $(0, +\infty)$ because $||f_n||=sup_{x \in (0,+\infty)f_n(x)= +\infty}$ as $\lim_{x \to 0}f_n(x)=+\infty$; the exchange of order between the series and the integral can thus be done in every interval of the kind $[M,+\infty)$; how can I justify it over the whole domain? – Lonidard Dec 20 '15 at 15:22
• @bharb It comes from $\frac1x$ factor in each $f_n(x)$. If you consider instead $xf(x)$ this problem will disappear. – Start wearing purple Dec 20 '15 at 15:26