I am required to find the limits of two "siblings" using the same idea, they are: $$\lim_{n\to\infty}\sin{\left(\frac{1}{pn+r}\right)}\sum_{k=1}^{n}\sin{\left(\frac{2pk-2p+r}{2pn}\right)}$$ with $0<r<2p$ and $$\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^{n}k \arctan{\left(\frac{pk-p+1}{pn}\right)}$$ where $p>1$.
For the first I tried $\sum_{k=1}^{n}sin(\frac{2pk+r}{2pn}-\frac{1}{n})=cos(\frac{1}{n})\sum_{k=1}^{n}sin(\frac{2pk+r}{2pn})-sin(\frac{1}{n})\sum_{k=1}^{n}cos(\frac{2pk+r}{2pn})$ but I don`t know how to find the limit using the ,,squeeze" theorem. Well, for the second I stumble upon the same problem as to the first: $\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=1}^{n}\frac{k}{n}\arctan{\left(\frac{k}{n}+\frac{p-1}{pn}\right)}$. Is this converging to $\int_{0}^{1}x\arctan{(x)}dx$ as $n\to\infty$? Is this the key ideea?
Hint. As regards the second sum, note that, $$0\leq \frac{k-1}{n}\leq \frac{pk-p+1}{pn}\leq \frac{k}{n}\leq 1,$$ for $1\leq k\leq n$ and $p>1$, and it follows that $$\frac{1}{n}\sum_{k=1}^{n}\frac{k}{n}\arctan{\big(\frac{k-1}{n}\big)}\leq \frac{1}{n^2}\sum_{k=1}^{n}k\arctan{\big(\frac{pk-p+1}{pn}\big)}\leq \frac{1}{n}\sum_{k=1}^{n}\frac{k}{n}\arctan{\big(\frac{k}{n}\big)}$$ The left side can be written as $$\frac{1}{n}\cdot\frac{1}{n}\sum_{k=0}^{n-1}\arctan{\big(\frac{k}{n}\big)}+ \frac{1}{n}\sum_{k=0}^{n-1}\frac{k}{n}\arctan{\big(\frac{k}{n}\big)}.$$ Use a similar approach for the first sum. Since $0<r<2p$, we have that $$0\leq \frac{k-1}{n}\leq \frac{2pk-2p+r}{2pn}\leq \frac{k}{n}\leq 1$$ for $1\leq k\leq n-1$. Hence $$\sum_{k=1}^{n}\sin{\left(\frac{k-1}{n}\right)}\leq \sum_{k=1}^{n}\sin{\left(\frac{2pk-2p+r}{2pn}\right)}\leq \sum_{k=1}^{n}\sin{\left(\frac{k}{n}\right)}.$$ Can you take it from here?