Science:Math Exam Resources/Courses/MATH101/April 2016/Question 06 (a)/Solution 1

The $N$th partial sum is

${\begin{aligned}s_{N}=\sum _{n=3}^{N}{\bigg (}\!\cos {\Big (}{\frac {\pi }{n}}{\Big )}-\cos {\Big (}{\frac {\pi }{n+1}}{\Big )}{\bigg )}&={\bigg (}\!\cos {\Big (}{\frac {\pi }{3}}{\Big )}{\color {OrangeRed}-\cos {\Big (}{\frac {\pi }{4}}{\Big )}}{\bigg )}+{\bigg (}\!{\color {OrangeRed}\cos {\Big (}{\frac {\pi }{4}}{\Big )}}{\color {OliveGreen}-\cos {\Big (}{\frac {\pi }{5}}{\Big )}}{\bigg )}+\cdots +{\bigg (}\!{\color {Maroon}\cos {\Big (}{\frac {\pi }{N}}}{\Big )}-\cos {\Big (}{\frac {\pi }{N+1}}{\Big )}{\bigg )}\\&=\cos {\Big (}{\frac {\pi }{3}}{\Big )}-\cos {\Big (}{\frac {\pi }{N+1}}{\Big )}.\end{aligned}}$

As $N\to \infty$, the argument ${\frac {\pi }{N+1}}$ converges to $0$, and $\cos x$ is continuous at $x=0$.

Thus the value of the series is $\lim _{N\to \infty }s_{N}=\lim _{N\to \infty }{\bigg (}\cos {\Big (}{\frac {\pi }{3}}{\Big )}-\cos {\Big (}{\frac {\pi }{N+1}}{\Big )}{\bigg )}=\cos {\Big (}{\frac {\pi }{3}}{\Big )}-\cos {\Big (}\lim _{N\to \infty }{\frac {\pi }{N+1}}{\Big )}=\cos {\Big (}{\frac {\pi }{3}}{\Big )}-\cos(0)={\color {blue}-{\frac {1}{2}}}.$

Therefore, the answer is $-{\frac {1}{2}}.$