What is $\sum^\infty_{i=1}\frac1{\prod^n_{j=0}i+jx}=\sum^\infty_{i=0}\frac1{i(i+x)(i+2x)\cdots(i+nx)}$? (This is a follow up to This question. Note that the linked question is about the case $n=1$.)
What is $$\sum^\infty_{i=1}\frac1{\prod^n_{j=0}i+jx}=\sum^\infty_{i=1}\frac1{i(i+x)(i+2x)\cdots(i+nx)}$$
My attempt:
I tried doing something similar to what J.G. did, by rewriting the sum as an integral.
We first note that $\int^1_0 t^{n-1}dt=\frac{t^n}{n}\big|^1_0=\frac{1^n-0^n}{n}=\frac 1n$. Then:
$$\sum^\infty_{i=1}\frac1{\prod^n_{j=0}i+jx}$$
$$=\sum^\infty_{i=1}\prod^n_{j=0}\frac1{i+jx}$$
$$=\sum^\infty_{i=1}\prod^n_{j=0}\int^1_0 t^{i+jx-1}dt$$
$$=\int^1_0 \prod^n_{j=0}t^{jx}\sum^\infty_{i=1}t^{i-1}dt$$
$$=\int^1_0t^{x\sum^n_{j=0}j}\frac{1}{1-t}dt$$
$$=\int^1_0\frac{t^{\frac{xn(n+1)}2}}{1-t}dt$$
However, when I plug in $n=1$, this gives me $$\int^1_0\frac{t^x}{1-t}dt$$
What I should have gotten is $\frac1x\int^1_0\frac{1-t^x}{1-t}dt$, which means I did something wrong. Where is my mistake, and how should I have done it? (The transition from $=\sum^\infty_{i=0}\prod^n_{j=0}\int^1_0 t^{i+jx-1}dt$ to $=\int^1_0 \prod^n_{j=0}t^{jx}\sum^\infty_{i=0}t^{i-1}dt$ seems questionable is incorrect as GEdgar pointed out it is incorrect to swap the product and sum or integral signs.)
 A: In general we have
$\sum_{i=1}^\infty \frac{1}{i (i+a_1)\cdots(i+a_n)}=(-1)^{n-1}\sum_{k=1}^n \frac{ H_{a_k}}{a_k \prod_{1\le j\le n,j \not= k}(a_k-a_j)}$
due to partial fraction expansion of LHS's summand and that $\sum _{n=1}^{\infty } \frac{1}{(a+n) (b+n)}=\frac{H_a-H_b}{a-b}$, thus OP's sum equals
$$(-1)^{n-1}x^{-n}\sum_{k=1}^n \frac{ H_{kx}}{k \prod_{1\le j\le n,j \not= k}(k-j)}$$
A: Yes the interchange is not valid as products are involved.
Also at $i=0$ there is a divergence thus we will ignore $i=0$
You can start by using partial fraction on the product to get:
$$\prod^n_{k=0}\frac{1\:}{m+kx\:}= \frac{a_0}{m} +\sum _{k=1}^n\frac{a_k}{m+kx}$$
To determine $a_j :j \not=0 $ we integrate both sides around circular contour centred at $\frac{-m}  {j}$ with radius $\xi$ as $\xi\rightarrow 0$
Therefore:
$$\lim _{\xi \to 0}\int _0^{2\pi }i\xi \:e^{i\theta \:}d\theta \:\prod ^n_{k=0}\frac{1}{m+k\left(-\frac{m}{j}+\xi \:e^{i\theta \:}\right)\:}=\frac{2i\pi a_j}{j}$$
$$\frac{2i\pi a_j}{j}=\lim _{\xi \to 0}\int _0^{2\pi }i\xi \:e^{i\theta \:}d\theta\:\prod ^n_{k=0}\frac{1}{m+k\left(-\frac{m}{j}+\xi \:e^{i\theta \:}\right)\:}$$
$$\frac{2i\pi a_j}{j}=\lim _{\xi \to 0}\int _0^{2\pi }id\theta\:\prod ^n_{k=0\\ k\not=j\ 
}\frac{1}{m+k\left(-\frac{m}{j}+\xi \:e^{i\theta \:}\right)\:}$$
Therefore:
$$a_j=j\prod ^n_{k=0\\ k\not=j\ 
}\frac{j}{mj-km}=j\:\left(\frac{j}{m}\right)^n\frac{\left(-1\right)^{n-j}}{j!\left(n-j\right)!}$$
Now plugging everything in sum we get:
$$\sum _{m=1}^{\infty }\left(\sum _{j=0}^nj\left(\left(\frac{j}{m}\right)^n\frac{\left(-1\right)^{n-j}}{n!}\begin{pmatrix}n\\ j\end{pmatrix}\right)\frac{1}{m+jx}\right)=S$$
Interchaging order of summation and use $$\int _0^1\:u^{m+xj-1}du=\frac{1}{m+xj}$$ We get:
$$S=\sum _{j=0}^n j\left(\frac{j^n\left(-1\right)^{n-j}}{n!}\begin{pmatrix}n\\ \:\:j\end{pmatrix}\right)\int _0^1\:\sum _{m=1}^{\infty }\left(\frac{u^m}{m^n}\right)\frac{u^{jx}}{u}\:\:du$$
$$S=\sum _{j=0}^n j\left(\frac{j^n\left(-1\right)^{n-j}}{n!}\begin{pmatrix}n\\ \:j\end{pmatrix}\right)\int _0^1 \text{Li}_n\left(u\right)\frac{u^{jx}}{u}\:\:du$$
This is progress so far, the final task is to evaluate sum then solve integral and we will be done. Unfortunatly I can't find closed form of series in meantime . When I solve it I'll update you.
Update:
We can write sum $S$ as :
$$\frac{d^{n+1}}{dx^{n+1}}\left(\int _0^1\:\sum _{j=0}^n \:\left(\frac{\left(-1\right)^{n-j}u^{jx}}{n!}\begin{pmatrix}n\\ \:j\end{pmatrix}\right)\frac{\text{Li}_n\left(u\right)}{u\:\ln ^{n+1}\left(u\right)}\:\:du\right)\:=\frac{d^{n+1}}{dx^{n+1}}\left(\int _0^1\frac{1}{n!}\:\left(u^x-1\right)^n\frac{\text{Li}_n\left(u\right)}{u\left(\ln \left(u\right)\right)^{n+1}}\:\:du\right)\:$$
