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$$f(x)=\sum_{k=1}^\infty\frac{x^kk!}{(x+1)(2x+1)\dots(kx+1)}$$

I'm asked to investigate this series and determine where it's continuous, but I have no idea where to even start with this.

I was able to determine that the series converges when 0<x<1, and diverges when x$\ge$1, using the raabe test, but after this I don't know how to to show continuity anywhere.

EDIT: I determined the convergence as follows:

$$\rho_k\equiv k\left(\frac{x^kk!}{(x+1)(2x+1)\dots(kx+1)}\frac{(x+1)\dots(kx+1)((k+1)x+1)}{x^{k+1}(k+1)!}-1\right)$$

$$\rho_k=k\left(\frac{(k+1)x+1}{(k+1)x}-1\right)=\frac{k}{(k+1)x}$$

$$\lim_{k\to\infty}\rho_k=\frac{1}{x}\lim_{k\to\infty}\frac{k}{k+1}=\frac{1}{x}$$

So the series converges when $\frac{1}{x}>1$ and diverges when $\frac{1}{x}<1$, so $x<1$ and $x>1$. When $x<0$ we have $\frac{1}{x}>1\Rightarrow 1<x$ which is a contradiction, so $x\nless0$

When $x=1$ we have

$$f(1)=\sum_{k=1}^\infty\frac{1^kk!}{(1+1)(2\cdot 1+1)\dots(k\cdot 1+1)}=\sum_{k=1}^\infty\frac{k!}{(k+1)!}=\sum_{k=1}^\infty\frac{1}{k+1}$$ which diverges

So we have $x\ge 1$ diverges and $0\le x<1$ converges

Andrew Lys
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    Actually this converges iff $0\leqslant x<1$, and in this case $f(x)=x/(1-x)$. – metamorphy Dec 02 '22 at 06:24
  • Hi :) i don't know, if it helps, but maybe you can use binomial coefficients. We have $\displaystyle (x+1)\ldots(kx+1)=x^k\cdot (1+\frac 1x)\ldots (k+\frac 1x)=x^k\cdot \binom{k+\frac 1x}{k}\cdot k!$. Hence your function can be written as $\displaystyle f(x)=\sum_{k=1}^\infty\frac{1}{\binom{k+\frac 1x}{k}}$. – Jochen Dec 02 '22 at 07:18
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    @metamorphy I posted my work for why it converges as I said. It would help if you could look over it and point out the mistake. Also, why would $f(x)=\frac{x}{1-x}$? I dont see how you got that – Andrew Lys Dec 02 '22 at 20:25
  • @Jochen That's a pretty cool identity, but I have even less of an idea of how to manipulate binomial coefficients to somehow extract uniform convergence on some domain. – Andrew Lys Dec 02 '22 at 20:54
  • @AndrewLys You wrote "the series converges when $\frac1x > 1$" which is correct. Then you wrote "$x\leq 1$ diverges." Those two statements contradict each other. – irchans Dec 02 '22 at 21:59
  • @irschans, I see – Andrew Lys Dec 02 '22 at 22:10

2 Answers2

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As already established, the given series $$f(x)=\sum_{k=1}^\infty f_k(x),\qquad f_k(x)=\frac{x^k\ k!}{(x+1)\cdots(kx+1)}$$ converges if and only if $0\leqslant x<1$.

Moreover, it converges uniformly on $x\in[0,a]$ for any $0<a<1$ (by Weierstrass M-test, since $0\leqslant f_k(x)\leqslant f_k(a)$ for $0\leqslant x\leqslant a$), hence $f(x)$ is continuous (as each $f_k$ is).

It is not asked for, but the equality $\color{blue}{f(x)=x/(1-x)}$ follows from $$\sum_{k=1}^n f_k(x)=\frac{x}{1-x}\big(1-(n+1)f_n(x)\big),$$ which is easy to show using induction on $n$.

metamorphy
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Manipulating the numerator and denominator of the sum slightly, it turns out that the infinite series can be expressed in terms of the hypergeometric function

$$f(x)=\sum_{k=0}^\infty\frac{(k+1)!}{(1+1/x)(2+1/x)\dots(k+1+1/x)}\\=\frac{x}{1+x}\sum_{k=0}^\infty\frac{(1)_k(2)_k}{(2+1/x)_k}\\=\frac{x}{1+x}{}_2F_1(1,2,2+1/x;1)$$

From considerations contained in this question, it can be shown that ${}_2F_1(1,2,2+1/x,z)$ is uniformly continuous as long as $|z|<1$. This leaves the present problem as an edge case, which has to be treated separately, due to the singularity present at $x=z=1$.

To assess continuity, we look at the integral representation

$$B(b,c-b){}_2F_1(a,b,c;1)=\int_0^1 x^{b-1}(1-x)^{c-b-a-1}$$

which from the definition of the beta integral gives,

$$f(x)=\frac{x}{1+x}\frac{B(2,1/x-1)}{B(2,1/x)}=\frac{x}{1-x}~~,~~ |x|<1$$

as @metamorphy pointed out in the comments. This is obviously uniformly continuous in any subset of the unit disk (not including its boundary).

DinosaurEgg
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