Derivatives of Infinite Product that Diverges to 0

Question

Consider the function given by $$f_n(x) = \prod\limits_{k=2}^n \left( 1-\frac{x}{k} \right).$$ Then for all $x \in (0,2)$, we have that $$\lim_{n\to\infty}\; f_n(x) = \prod\limits_{k=2}^\infty \left( 1-\frac{x}{k} \right) =0. $$ This follows from the fact that if $q_k \in [0,1)$ for all $k$, then $\prod\limits_{k=1}^{\infty} (1-q_k) = 0$ if and only if $\sum\limits_{k=1}^\infty q_k$ diverges (See for example here and here).

I am interested in the derivatives of this function $f_n$. For instance, let $f_n^{(j)}$ denote the $j$-th derivative of $f_n$. Is it true that $\lim_{n\to\infty} f_n^{(j)}(x) = 0$ for all $x\in(0,2)$?

Considering the first derivative, we have that from the product rule: $$ f_n'(x) = \sum\limits_{k=2}^n \frac{-1}{k} \prod\limits_{i \neq k} \left(1-\frac{x}{i} \right)\tag1 \label 1 .$$ Now, we have that \begin{align}\prod\limits_{i \neq k} \left(1-\frac{x}{i} \right) &\leq \prod\limits_{i =2}^{n-1} \left(1-\frac{x}{i} \right)\\ &= \exp\left\{ \sum_{i=2}^{n-1} \log\left(1-\frac{x}{i} \right) \right\} \\ &\leq \exp\left\{ \sum_{i=2}^{n-1} -\frac{x}{i} \right\} \tag2 \\ &\leq \exp\left\{-x\log(n)+x \right\}\tag3\\ &= {\left(\frac{e}{n}\right)}^x.\end{align} Where (2) above follows from the fact that $\log(1-y)\leq -y$ for all $y<1$, and (3) follows from $\sum\limits_{i=2}^{n-1}\frac{1}{i} \geq \log(n)-1$. Now, by noticing all the terms in the sum in $\ref1$ are nonpositive, we have the following \begin{align*} f_n'(x) \geq {\left(\frac{e}{n}\right)}^x\sum\limits_{k=2}^n \frac{-1}{k} \geq -{\left(\frac{e}{n}\right)}^x \log(n) \end{align*} Finally, because $f_n'(x)\leq 0$ and $\lim_{n \to \infty} {\left(\frac{e}{n}\right)}^x \log(n) = 0$, we are able to conclude that $\lim \; f_n'(x) = 0$ for all $x \in (0,2)$.

Now, is there a way to show this for the $j$-th derivative $f_n^{(j)}$? I am unable to even currently show that the second derivative is zero (in my simulations it seems to approach zero, but at a rate slower than the first derivative). An idea that I have briefly considered using is considering the differentiation of the infinite product (see here and similarly here). However in my case my infinite product diverges to zero, so I am not sure how useful this will be. Also, I am interested in the higher order derivatives of this infinite product, not just the first derivative.

I appreciate any input on this problem!

Answer 1

"The shortest path to truth in $\Bbb R$ is through $\Bbb C.$" -- Hadamard.

We can use a powerful but basic result from complex analysis to quickly answer this: Extend the formula for each $f_n$ to the domain $D=\{z\in \Bbb C: |z-1|<1\}\cup \{z\in \Bbb C: |z-2|<1\}.$

Then each $f_n$ is analytic on $D$ and the sequence $(f_n)_{n\geq 2}$ converges uniformly to $0$ on any $E\subset D$ such that $E$ is a closed subset of $\Bbb C.$ For $x\in D$ take $r_x>0$ such that $\{z: |z-x|\leq r_x\}\subset D.$

For any $j\geq 0$ we have $$f_n^{(j)}(x)=\frac {j!}{2\pi i}\int_{|z-x|=r_x}\frac {f_n(z)}{(z-x)^{j+1}}dz.$$ For fixed $j$ we have $|f_n(z)|\to 0$ uniformly on $\{z:|z-x|=r_x\}$ as $n\to \infty,$ so $f_n^{(j)}(x)\to 0$ as $n\to \infty.$

Moreover, as $n\to\infty,$ for each $j\geq 0$ the sequence $(\;f_n^{(j)}\;)_{n\geq 2}$ converges uniformly to $0$ on $[s,2]$ for any $s>0.$

This may be outside the scope of your background but it shows the strength of complex analysis applied to real analysis.

Answer 2

as a HINT, notice that $$ \eqalign{ & f(x,n) = \prod\limits_{2\, \le \,k\, \le \;n} {\left( {1 - {x \over k}} \right)} = {{\left( { - 1} \right)^{\,n - 1} \prod\limits_{2\, \le \,k\, \le \;n} {\left( {x - k} \right)} } \over {\prod\limits_{2\, \le \,k\, \le \;n} k }} = \cr & = {{\left( { - 1} \right)^{\,n - 1} \prod\limits_{0\, \le \,k\, \le \;n - 2} {\left( {x - 2 - k} \right)} } \over {n!}} = {{\left( { - 1} \right)^{\,n - 1} \left( {x - 2} \right)^{\,\underline {\,n - 1\,} } } \over {n!}} = \cr & = {{\left( { - 1} \right)^{\,n - 1} } \over {x - 1}}\left( \matrix{ x - 1 \cr n \cr} \right) = \left( { - 1} \right)^{\,n - 1} {{\Gamma \left( {x - 1} \right)} \over {\Gamma \left( {x - n} \right)\Gamma \left( {n + 1} \right)}} \cr} $$ and also $$ \eqalign{ & f(x,n) = {{\left( { - 1} \right)^{\,n - 1} } \over {x - 1}}\left( \matrix{ x - 1 \cr n \cr} \right) = {1 \over {1 - x}}\left( \matrix{ n - x \cr n \cr} \right) = {1 \over {1 - x}}\left( \matrix{ n - x \cr - x \cr} \right) = \cr & = {1 \over {1 - x}}{{\Gamma (n - x + 1)} \over {\Gamma (n + 1)\Gamma (1 - x)}} = {{\Gamma (n + 1 - x)} \over {\Gamma (n + 1)\Gamma (2 - x)}} \cr} $$