I'm working through exercise 19 in section 1.2.9 of Knuth's The Art of Computer Programming volume 1, which asks, using the definition
$$ H_{p/q} = \sum_{n \geq 1}\left(\frac{1}{n} - \frac{1}{n+p/q}\right)\text{,} $$
to prove that
$$ H_{p/q} = \frac{q}{p} - \frac{\pi}{2}\cot\frac{p}{q}\pi - \ln 2q + 2\sum_{0 < k < q/2}\cos\frac{2pk}{q}\pi \cdot \ln\sin\frac{k}{q}\pi\text{,} $$
when $p$ and $q$ are integers with $0 < p < q$. The desired solution starts by using Abel's limit theorem to find
$$ \begin{aligned} H_{p/q} &= \sum_{n \geq 1}\left(\frac{1}{n} - \frac{1}{n+p/q}\right)\\ &= \lim_{x \to 1-}\sum_{n \geq 1}\left(\frac{1}{n} - \frac{1}{n + p/q}\right)x^{p+nq} & \text{(1)}\\ &= \lim_{x \to 1-}\left(\sum_{k=0}^{q-1}\omega^{-kp}\ln(1-\omega^k x) - x^p\ln(1-x^q) + \frac{q}{p}x^p\right)\text{,} & \text{(2)}\\ \end{aligned} $$
where $\omega = e^{2\pi i/q}$.
I'm having trouble getting completely from (1) to (2). I'm able to derive the first two terms of (2), but am unsure of the origin of the $\frac{q}{p}x^p$ term.
So far I have, in the limit as $x \to 1-$,
$$ \begin{aligned} &\sum_{n \geq 1}\left(\frac{1}{n} - \frac{1}{n + p/q}\right)x^{p+nq}\\ &\quad= -\sum_{n \geq 1}\frac{1}{n + p/q}x^{p+nq} + \sum_{n \geq 1}\frac{1}{n}x^{p+nq}\\ &\quad= -\sum_{\substack{n > 0\\p + nq \bmod q = p}}\frac{q}{p + nq}x^{p+nq} + \sum_{n \geq 1}\frac{1}{n}x^{p+nq}\\ &\quad= -\frac{1}{q}\sum_{0 \leq k < q}\omega^{-kp}\sum_{n \geq 1}\frac{q}{n}\omega^{kn}x^{n} + \sum_{n \geq 1}\frac{1}{n}x^{p+nq} & \text{by Eq. (13)}\\ &\quad= \sum_{0 \leq k < q}\omega^{-kp}\sum_{n \geq 1}\frac{(-1)^{n+1}}{n}(-\omega^{k}x)^{n} + \sum_{n \geq 1}\frac{1}{n}x^{p+nq}\\ &\quad= \sum_{0 \leq k < q}\omega^{-kp}\ln(1 - \omega^{k}x) + \sum_{n \geq 1}\frac{1}{n}x^{p+nq} & \text{by Eq. (24)}\\ &\quad= \sum_{0 \leq k < q}\omega^{-kp}\ln(1 - \omega^{k}x) - x^{p}\sum_{n \geq 1}\frac{(-1)^{n+1}}{n}(-x^q)^{n}\\ &\quad= \sum_{0 \leq k < q}\omega^{-kp}\ln(1 - \omega^{k}x) - x^{p}\ln(1 - x^q) & \text{by Eq. (24)}\text{.}\\ \end{aligned} $$
Where is the $\frac{q}{p}x^{p}$ term supposed to come from?
Equations referenced include
$$ \sum_{\substack{n > 0\\n \bmod m = r}}a_{n}z^{n} = \frac{1}{m}\sum_{0 \leq k < m}\omega^{-kr}G(\omega^{k}z)\text{,}\quad\text{$0 \leq r < m$,} \tag{13} $$
with $\omega = e^{2\pi i/m}$ and $G(z)$ the generating function for $\langle a_n \rangle$; and
$$ \ln(1 + z) = \sum_{k \geq 1}\frac{(-1)^{k+1}}{k}z^{k}\text{.} \tag{24} $$
I've also looked through errata to see if this was an error in the printing I have—the twenty-fifth of the third edition—but to no avail. Thanks for any insight!
Thanks, Mark. It turns out I was missing a term from the application of Eq. (13), which apparently is not for $n > 0$ but for $n \geq 0$. That is,
$$ \begin{aligned} &-\sum_{\substack{n > 0\\p+nq \bmod q = p}}\frac{q}{p+nq}x^{p+nq}\\ &\quad= \frac{q}{p}x^{p} - \sum_{\substack{n \geq 0\\p+nq \bmod q = p}}\frac{q}{p+nq}x^{p+nq}\\ &\quad= \frac{q}{p}x^{p} - \frac{1}{q}\sum_{0 \leq k < q}\omega^{-kp}\sum_{n \geq 1}\frac{q}{n}\omega^{kn}x^{n}\text{,} & \quad \text{by Eq. (13)}\\ \end{aligned} $$
meaning the referenced equation should really read
$$ \sum_{\substack{n \geq 0\\n \bmod m = r}}a_{n}z^{n} = \frac{1}{m}\sum_{0 \leq k < m}\omega^{-kr}G(\omega^{k}z)\text{,}\quad\text{$0 \leq r < m$.} \tag{13} $$
(Note that this is a known erratum in the twenty-fifth printing.)