Proving $\lim_{n\to\infty}\sum_{k=qn}^{pn}\frac1k=\log\frac{p}{q}$ This is a question from Apostol Vol 1, it states:

If $p$ and $q$ are fixed integers, $p≥q≥1$, show that $$\lim_{n\to\infty}\sum_{k=qn}^{pn}\frac1k=\log\frac{p}{q}$$

It seems likely to me that this sum must be somehow connected to the integral $\int_q^p\frac1xdx$. I have tried to convert the sum so that it adopts the typical form of a Riemann sum, but I have been unable to. Another possibility that I thought of was using the integral estimate for infinite series, but again, this hasn't yielded anything. How should I approach the problem?
 A: Note that the sum under limit has each term tending to $0$ and hence we can ignore the first term $1/qn$ and replace the sum with $$\sum_{k=qn+1}^{pn}\frac{1}{k}$$ which can be further rewritten as $$\sum_{i=q}^{p-1}\sum_{j=1}^{n}\frac{1}{ni+j}=\sum_{i=q}^{p-1}\frac{1}{n}\sum_{j=1}^{n}\frac{1}{i+(j/n)}$$ Just focus on the inner sum which is a typical Riemann sum and the desired limit is thus $$\sum_{i=q} ^{p-1}\int_{0}^{1}\frac{dx}{x+i}$$ Putting $x+i=t$ we get $$\sum_{i=q} ^{p-1}\int_{i}^{i+1}\frac{dt} {t} =\int_{q} ^{p} \frac{dt} {t} =\log\frac{p} {q} $$

Another way to look at your sum is to recognize that it has $pn-qn=N$ terms (ignoring first term) and thus we need a partition involving those many terms. Rewrite this as $$\sum_{k=1}^{N}\frac{1}{qn+k}=\sum_{k=1}^{N}\frac{1}{qN/(p-q)+k}$$ and this is further expressed as $$\frac{p-q} {N}\sum_{k=1}^{N}\dfrac{1}{q+k\cdot\dfrac{p-q}{N}}$$ and as $n\to\infty $ we have $N\to\infty $ so the desired limit is $\int_{q}^{p} \frac{dx} {x} $. 
A: We have
$$\lim_{n\to\infty}\sum^{pn}_{k=qn}\frac1k=\lim_{n\to\infty}\sum^{pn}_{k=qn}\frac1{\frac kn}\frac1n$$
This is in the form of a Riemann sum, so the limit is given by
$$\lim_{n\to\infty}\sum^{pn}_{k=qn}\frac1{\frac kn}\frac1n=\int_q^p\frac 1x\mathrm dx=\log\left(\frac pq\right)$$
A: Without Riemann sum.
$$\sum_{k=a}^{b}\frac1k=H_b-H_{a-1}$$
$$\sum_{k=q\,n}^{p\,n}\frac1k=H_{pn}-H_{qn-1}$$
Now, using the asymptotics of harmonic numbers
$$H_r=\log (r)+\gamma +\frac{1}{2 r}+O\left(\frac{1}{r^2}\right)$$
$$\sum_{k=q\,n}^{p\,n}\frac1k=\log \left(\frac{p}{q}\right)+\frac{p+q}{2  p q}\frac 1n+O\left(\frac{1}{n^2}\right)$$ which shows the limit and also how it is approached.
A: Hint:
$$\sum_{k=nq}^{np}\frac{1}{k} > \int_{pq}^{np+1}\frac{dx}{x}.$$
(Recall this is the same kind of thing you do in calculus when deriving the integral test.) You can get an upper bound the same way.
