How do I use the following difference root (LDE) to extract a partial sum of the harmonic series below? I asked WolframAlpha about some partial sum formula for the following divergent series:
$$\sum_{x=2}^\infty \frac{H_x}{ax\pm1} = ??$$
Where $H_x$ is the $x$th harmonic number.
It informed me that it is indeed divergent, then spat out the following for the partial sum formula:
$$DifferenceRoot[\{y, n\}, Function \{(n + 1) (a n \pm 1) y(n) - (n + 1) (2 a n + a \pm 2) y(n + 1) + (n + 1) (a n + a \pm 1) y(n + 2) - 1 = 0, y (0) = 0, y (1) = 0\}](m + 1) \mp \frac{1}{1 \pm a}$$
What am I supposed to do with that? I checked the documentation, it's not entirely clear how you can use this without a Wolfram Notebook product license. What techniques should I even be thinking about here to solve this, something regarding linear differential equations?

You have a problem because of the $\pm 1$. If it was just
$$\sum_{n=2}^p \frac {H_n}{an}=\frac{6 \left(H_p\right){}^2-6 \psi ^{(1)}(p+1)+\pi ^2-12}{12 a}$$ which is an upper or lower bound for your summation.
Assuming that $a$ is large, what you could do is to write
$$\sum_{n=2}^p \frac {H_n}{an\pm 1}\sim \sum_{n=2}^m \frac {H_n}{an\pm 1}+\sum_{n=m+1}^p \frac {H_n}{an}$$ Compute the first summation and the second one is
$$\sum_{n=m+1}^p \frac {H_n}{an}=\frac{\left(H_p\right){}^2-\left(H_m\right){}^2+\psi ^{(1)}(m+1)-\psi^{(1)}(p+1)}{2 a}$$
Edit
For the case of
$$S_k=\sum_{n=2}^p \frac {H_n}{n+k}$$ where $k$ is an integer, the formulae are quite simple
$$S_1=\frac{1}{2} \left(\left(H_{p+1}\right){}^2-H_{p+1}^{(2)}-1\right)$$
$$S_2=\frac{1}{6} \left(3 \left(H_{p+2}\right){}^2-3 H_{p+2}^{(2)}+\frac{6}{p+2}-8\right)$$
$$S_3=\frac{1}{2} \left(\left(H_{p+3}\right){}^2+\frac{1}{p+2}+\frac{3}{p+3}+\psi^{(1)}(p+4)-\frac{\pi ^2}{6}-4\right)$$
