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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.

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  • $\begingroup$ I'm not sure if this is a mathematically sound step, but after $\lim_{x\to\infty}\sum_{k=1}^\infty \frac{(-1)^{k-1}}{kx^{k-1}} \sum_{n=1}^\infty\frac{1}{n^k(2n-1)^k}$, the inner sum can be computed exactly using partial fractions and the zeta function. $\endgroup$ Mar 28 '20 at 1:59
  • $\begingroup$ @VVejalla I verified the double sum numerically. You changed the upper limit of the inner sum. $\endgroup$ Mar 28 '20 at 2:03
  • $\begingroup$ Yeah, I substituted $\infty$ for $x$. That's why I'm not sure if it is mathematically sound. $\endgroup$ Mar 28 '20 at 2:04
  • $\begingroup$ @VVejalla we cant do that because there is $x$ in the summand $\endgroup$ Mar 28 '20 at 2:05
  • $\begingroup$ That's what I thought too. What happened with the $\mp$ in the double sum in the question? $\endgroup$ Mar 28 '20 at 2:07
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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

$$\psi(z+1)=-\gamma+\sum_{n=1}^{\infty}\frac{z}{n(n+z)}$$

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  • $\begingroup$ Beautiful (+1). $\endgroup$ Mar 28 '20 at 2:26
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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,

$$\lim_{m\to \infty } \, m \log (p)=\psi ^{(0)}\left(1+\frac{1}{a}\right)+\gamma = H_{\frac{1}{a}}$$

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  • $\begingroup$ Nice approach Wolf +1 $\endgroup$ Apr 2 '20 at 2:54

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