Limit $\lim\limits_{n \to \infty} \sum\limits_{k = 0}^n \dfrac{1}{qn+kp+1}$ I have the following limit of a sum:
$$\lim\limits_{n \to \infty} \sum\limits_{k = 0}^n \dfrac{1}{qn+kp+1}$$
Where $p \in \mathbb{N} \setminus \{0, 1\}$ and $q > 0$.
I am trying to convert this limit into an integral by recognizing it as a Riemann sum. I'm fairly new to this concept of converting from limit to integral so I'm pretty lost. Anyway, this is what I tried:
$$\lim\limits_{n \to \infty} \sum\limits_{k = 0}^n \dfrac{1}{qn+kp+1} = \lim\limits_{n \to \infty} \dfrac{1}{n} \sum\limits_{k=0}^n \dfrac{1}{q+ p\frac{k}{n}+ \frac{1}{n}}$$
I am thinking of using this:
$$\int_a^b f(x) dx = \lim\limits_{n \to \infty} \sum_{k=1}^n f(a + k \cdot \Delta x) \cdot \Delta x$$
Where $\Delta x = \dfrac{b-a}{n}$. I think I would know how to do this if it wasn't for that $\dfrac{1}{n}$ term in the denominator of the sum's terms. If it wasn't for that $\dfrac{1}{n}$, I think the function I would have to use in the definite integral would be something like:
$$f(x) = \dfrac{1}{q+px}$$
But that $\dfrac{1}{n}$ really confuses me. What happens to it and how does that influence the final integral? And what is that integral? Also, if anyone knows of any resource from where I could learn more about this conversion of Riemann sum $\rightarrow$ definite integral (preferably with examples), I would really appreciate it if you could link me to those resources.
 A: You are correct, the $\frac{1}{n}$ term makes the problem more complicated. However, it does not contribute to the limit in the sense that removing it will not change the limit. The following solution uses that. 
Let $$S(p,q)=\lim_{n\to \infty} \sum_{k=0}^n \frac{1}{qn+pk+1},\qquad T(p,q)=\lim_{n\to \infty}\sum_{k=0}^n \frac{1}{qn+pk}$$
Claim 1: $T(p,q)=\int_0^1\frac{dx}{q+p x}$ for any $p,q>0$. 
Claim 1 can be proved using Riemann sums 
$$T(p,q)=\lim_{n\to \infty}\frac{1}{n} \sum_{k=0}^n \frac{1}{q+p\frac{k}{n}}.$$
Claim 2: $T(p,q)=S(p,q)$ for any $p,q>0$. To see this, notice that 
$$\frac{1}{qn+pk}-\frac{1}{qn+pk+1} = \frac{1}{(q n+p k) (qn+pk+1)}\le \frac{1}{q^2n^2}$$
So, 
$$\left|\sum_{k=0}^n \frac{1}{qn+pk+1}-\sum_{k=0}^n \frac{1}{qn+pk}\right|\le \frac{1}{q^2 n}$$ 
By taking $n\to \infty$, we can see that claim 2 holds. 
A: \begin{align*}
\sum_{k=0}^{n}\dfrac{1}{qn+p(k+1)}&=\dfrac{1}{p}\dfrac{p}{n}\sum_{k=0}^{n}\dfrac{1}{q+\dfrac{p(k+1)}{n}}\\
&=\dfrac{1}{p}\dfrac{p}{n}\sum_{k=0}^{n-1}\dfrac{1}{q+\dfrac{p(k+1)}{n}}+\dfrac{1}{p}\dfrac{p}{n}\dfrac{1}{q+\dfrac{p(n+1)}{n}},
\end{align*}
we have 
\begin{align*}
\lim_{n\rightarrow\infty}\dfrac{p}{n}\sum_{k=0}^{n-1}\dfrac{1}{q+\dfrac{p(k+1)}{n}}=\int_{0}^{p}\dfrac{1}{q+x}dx,
\end{align*}
the partition is $\left\{0,\dfrac{p}{n},\dfrac{2p}{n},\cdots,\dfrac{np}{n}\right\}$ with the right endpoints being selected.
On the other hand, 
\begin{align*}
\sum_{k=0}^{n}\dfrac{1}{qn+pk+1}&=\sum_{k=0}^{n}\dfrac{1}{qn+p(k+1)}\dfrac{qn+p(k+1)}{qn+pk+1},
\end{align*}
so
\begin{align*}
\sum_{k=0}^{n}\dfrac{1}{qn+pk+1}&\geq\sum_{k=0}^{n}\dfrac{1}{qn+p(k+1)},
\end{align*}
and
\begin{align*}
\sum_{k=0}^{n}\dfrac{1}{qn+pk+1}&=\sum_{k=0}^{n}\dfrac{1}{qn+p(k+1)}+\sum_{k=0}^{n}\dfrac{p-1}{(qn+p(k+1))(qn+pk+1)}\\
&\leq\sum_{k=0}^{n}\dfrac{1}{qn+p(k+1)}+\sum_{k=0}^{n}\dfrac{p-1}{q^{2}n^{2}}\\
&=\sum_{k=0}^{n}\dfrac{1}{qn+p(k+1)}+\dfrac{p-1}{q^{2}n},
\end{align*}
the rest follows by Squeeze Theorem.
