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


$\textbf{Part 2}$

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


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?

  • $\begingroup$ 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. $\endgroup$ – Colm Bhandal Dec 27 '15 at 13:54
  • $\begingroup$ @ColmBhandal Thank you for your observation. Got it fixed! $\endgroup$ – Lonidard Dec 27 '15 at 14:06
  • $\begingroup$ I think your derivation of $f(1)$ is off by $1$, you sum over $1/(n+1)!$, thus resulting in $f(1)=e-1$. $\endgroup$ – mlk Dec 28 '15 at 8:49
  • $\begingroup$ @mlk True. I'll fix it. $\endgroup$ – 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*}

  • $\begingroup$ 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! $\endgroup$ – Lonidard Dec 20 '15 at 15:16
  • $\begingroup$ @bharb I didn't know if we were allowed to use beta function. And if we weren't, then that formula could be shown by induction. $\endgroup$ – Start wearing purple Dec 20 '15 at 15:21
  • $\begingroup$ 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? $\endgroup$ – Lonidard Dec 20 '15 at 15:22
  • $\begingroup$ @bharb It comes from $\frac1x$ factor in each $f_n(x)$. If you consider instead $xf(x)$ this problem will disappear. $\endgroup$ – Start wearing purple Dec 20 '15 at 15:26

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.