# Evaluating iterated limits and double limit of $\frac{p}{q^2}\sum_{n=1}^q \operatorname{sin}\frac{n}{p}$

Investigate the existence of the two iterated limits and the double limit of the double sequence $f$ defined by

$$f(p,q)=\frac{p}{q^2}\sum_{n=1}^q \operatorname{sin}\frac{n}{p}$$

I think I need to reformulate the summation part to a simpler form but I don't know how. How can I investigate the limits of this double sequence? I would greatly appreciate some help.

$$f(p,q)=\frac{p}{q^2}\sum_{n=1}^q \operatorname{sin}\left(\frac{n}{p}\right)=$$

$$f(p,q)=\frac{p}{q^2} \left(\sin\left(\frac{1}{p}\right)+\sin\left(\frac{2}{p}\right)+\sin\left(\frac{3}{p}\right)+...+\sin\left(\frac{q-1}{p}\right)+\sin\left(\frac{q}{p}\right)\right)=$$

$$f(p,q)=\frac{p}{q^2}\left(\csc\left(\frac{1}{2p}\right)\sin\left(\frac{q}{2p}\right)\sin\left(\frac{q+1}{2p}\right)\right)=$$

$$f(p,q)=\frac{p\left(\csc\left(\frac{1}{2p}\right)\sin\left(\frac{q}{2p}\right)\sin\left(\frac{q+1}{2p}\right)\right)}{q^2}=$$

$$f(p,q)=\frac{p\csc\left(\frac{1}{2p}\right)\sin\left(\frac{q}{2p}\right)\sin\left(\frac{q+1}{2p}\right)}{q^2}$$

(for $\frac{1}{2\pi p}$ is not the element of Z -> the set of integers)

• Can you explain how the summation of the sin's become $csc(1/2p)sin(q/2p)sin(q+1/2p)$? – takecare May 13 '15 at 17:04
• Yes, you have a summation of sinus's so the most simple from of the summation is $\sum_{n=a}^q \sin(b)$ and that is equal to: $(-a+q+1)\sin(b)=-a\sin(b)+q\sin(b)+\sin(b)$ so if we transfrom it a bit: $\sum_{n=a}^q \sin(b/c)$ than we get: $(-a+q+1)\sin\left(\frac{b}{c}\right)=-a\sin\left(\frac{b}{c}\right)+q\sin\left(\frac{b}{c}\right)+\sin\left(\frac{b}{c}\right)$ – Jan May 13 '15 at 17:11
• I've it in a book called: Calculus - Early Transcendentals 6th Edition – Jan May 13 '15 at 17:21
• Thanks! Can you tell me which section of the book it's from? – takecare May 13 '15 at 17:24
• I've no idea, I have to study the complete book, so no idea I'm sorry! – Jan May 13 '15 at 17:26