Prove $\lim_{x\to\infty}\sum_{n=1}^x x\log\left(1+\frac1{xn(an+1)}\right)= H_{\frac1a}$ How to prove that 
$$\large\lim_{x\to\infty}\sum_{n=1}^x x\log\left(1+\frac1{xn(an+ 1)}\right)= H_{\frac1a}, \quad a\in \mathbb{R},\ |a|>1$$
This question is a formulated form of this problem.
where $H_r=\int_0^1\frac{1-z^r}{1-z}\ dz$ is the harmonic number. 
Any idea how to prove this identity? 
I tried to convert the log to integral but was not helpful, also I used the series expansion for the log and I got 
$$\lim_{x\to\infty}\sum_{n=1}^x x\log\left(1+\frac1{xn(an+ 1)}\right)=\lim_{x\to\infty}\sum_{n=1}^x\sum_{k=1}^\infty \frac{(-1)^{k-1}}{x^{k-1}k\ n^k(2n-1)^k}$$
and I have no idea how to continue with this double sum. Any help would be much appreciated. 
 A: We can use Tannery's theorem to take the limit of the summand, since it exists and also it is uniformly bounded,
$$x\log(1+\frac{1}{xn(an+b)})<\frac{1}{n(an+b)}$$
and since $\sum_{n=1}^{\infty}\frac{1}{n(an+b)}<\int_{1}^{\infty}\frac{dx}{x(ax+b)}=\frac{1}{b}\log(1+\frac{b}{a})$ exists and is finite,
then we compute that 
$$\lim_{x\to\infty}\sum_{n=1}^xx\log\Big(1+\frac{1}{xn(an+b)}\Big)=\sum_{n=1}^{\infty}\frac{1}{n(an+b)}=\frac{1}{b}(\psi(\frac{b}{a}+1)+\gamma)=\frac{1}{b}H_{\frac{b}{a}}$$
where $\psi$ is the digamma function.
Here we used the series representation of the digamma function
$$\psi(z+1)=-\gamma+\sum_{n=1}^{\infty}\frac{z}{n(n+z)}$$
A: The direct way.
Renaming $x\to m$ we can write
$$m \sum _{n=1}^m \log \left(\frac{1}{m n (a n+1)}+1\right)=m \log \left(\prod _{n=1}^m \left(\frac{1}{m n (a n+1)}+1\right)\right)$$
Now
$$p = \prod _{n=1}^m \left(\frac{1}{m n (a n+1)}+1\right)=\frac{\left(1-\frac{\sqrt{1-\frac{4 a}{m}}-1}{2 a}\right)_m \left(\frac{2 a+\sqrt{1-\frac{4 a}{m}}+1}{2 a}\right)_m}{\Gamma (m+1) \left(1+\frac{1}{a}\right)_m}$$
where $\left(k\right)_m=\frac{\Gamma(k+m)}{\Gamma(k)}$ is the Pochhammer symbol.
Using the asymptotic expansion of the gamma function we have to second order in $m$
$$p \overset{m\to\infty}\simeq \left(\frac{1}{12 a m^2}\right)\left(12 \gamma  a^2+12 a m^2+12 \gamma  a m+12 a (a+m+\gamma ) \psi ^{(0)}\left(1+\frac{1}{a}\right)\\+6 \gamma ^2 a-\pi ^2 a+6 a \psi ^{(0)}\left(1+\frac{1}{a}\right)^2\\-6 a \psi ^{(1)}\left(1+\frac{1}{a}\right)-12\right) + O(\frac{1}{m^2})$$
and, finally,
$$\lim_{m\to \infty } \, m \log (p)=\psi ^{(0)}\left(1+\frac{1}{a}\right)+\gamma = H_{\frac{1}{a}}$$
