Is this procedure on Harmonic Series correct? 
$\require{cancel}$I need to evaluate the following limit:
  $$\lim_{n\to\infty}\sum_{np}^{nq}\frac1k \quad \text{where $\quad p\geq1,q\geq1$}$$

The way I tackled this problem is as follows:
Recall the relation:
$$\sum_{k=1}^n\frac1k = \ln(n)+\gamma+o(1) \quad \text{as $n\to\infty$}$$
Therefore, assuming $q\geq p$, the initial limit can be expressed as:
$$\begin{align*}
\lim_{n\to\infty}\sum_{np}^{nq}\frac1k &= \lim_{n\to\infty}\left(\sum_{k=1}^{nq}\frac1k-\sum_{k=1}^{np-1}\frac1k\right)\\
&=\lim_{n\to\infty}\left(\ln(nq)+\cancel{\gamma}-\ln(np-1)\cancel{-\gamma}+o(1)\right)\\
&= \lim_{n\to\infty}\ln\left(\frac{nq}{np-1}+o(1)\right)=\ln\left(\lim_{n\to\infty}\left(\frac{\cancel{n}q}{\cancel{n}\left(p-\frac1n\right)}+o(1)\right)\right)=\boxed{\ln\left(\frac{q}{p}\right)}
\end{align*}$$

Are the result and procedure correct? Also, do I need to prove the initial identity (for a Homework Assignment)?

 A: The approach in the OP is fine.  

I thought it might be instructive to present another way forward that doesn't rely on the asymptotic expansion of the Harmonic number, but rather relies on only straightforward analysis and Riemann sums.  To the end, we proceed.

We simply write
$$\begin{align}
\sum_{k=np}^{nq}\frac{1}{k}&=\sum_{k=0}^{n(q-p)}\frac{1}{k+np}\\\\
&=\frac{1}{np}\sum_{k=0}^{n(q-p)}\frac{1}{1+\frac k{np}}\tag 1
\end{align}$$
We recognize that the right-hand side of $(1)$ is the Riemann sum for 
$$\begin{align}
\frac1p\int_0^{q-p}\frac{1}{1+x/p}\,dx&=\log\left(1+\frac{q-p}{p}\right)\\\\
&=\log(q/p)
\end{align}$$
Hence, we see that the coveted limit is 

$$\lim_{n\to \infty}\sum_{k=np}^{nq}\frac{1}{k}=\log(q/p)$$

as expected!
A: Going directly from
$\ln(x)
=\int_1^x \frac{dt}{t}
$,
we have
$\ln(k+1)-\ln(k)
=\int_k^{k+1} \frac{dt}{t}
$,
so
$\frac1{k+1}
< \ln(k+1)-\ln(k)
< \frac1{k}
$.
Summing,
$$\sum_{k=np}^{nq-1}\frac1{k+1}
<\sum_{k=np}^{nq-1} (\ln(k+1)-\ln(k))
<\sum_{k=np}^{nq-1} \frac1{k}
$$
so that
$$\sum_{k=np+1}^{nq}\frac1{k}
=\sum_{k=np}^{nq}\frac1{k}-\frac1{np}
< \ln(nq)-\ln(np)
=\ln(q/p)
<\sum_{k=np}^{nq-1} \frac1{k}
=\sum_{k=np}^{nq} \frac1{k}-\frac1{nq}
.$$
Therefore
$$-\frac1{np}
< \ln(q/p)-\sum_{k=np}^{nq}\frac1{k}
< -\frac1{nq}
$$
or
$$\frac1{nq}
< \sum_{k=np}^{nq}\frac1{k}-\ln(q/p)
< \frac1{np}
.$$
Now let $n \to \infty$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\lim_{n \to \infty}\sum_{np}^{nq}{1 \over k}:\ {\large ?}
\quad\mbox{where}\quad p \geq 1\,,\ q \geq 1}$.

\begin{align}
\lim_{n \to \infty}\sum_{np}^{nq}{1 \over k} & =
\lim_{n \to \infty}\pars{\sum_{k = 1}^{nq}{1 \over k} -
\sum_{k = 1}^{np - 1}{1 \over k}} =
\lim_{n \to \infty}\pars{H_{nq} - H_{np - 1}}\qquad
\pars{~H_{m}:\ Harmonic\ Number~}
\\[5mm] & =
\lim_{n \to \infty}\bracks{\Psi\pars{nq + 1} - \Psi\pars{np}}\qquad
\pars{~\Psi:\ Digamma\ Function~}
\\[5mm] & 
\lim_{n \to \infty}\bracks{\ln\pars{nq + 1} - \ln\pars{np}} =
\bbx{\ds{\ln\pars{q \over p}}}
\end{align}
