The N {\displaystyle N} th partial sum is
s N = ∑ n = 3 N ( cos ( π n ) − cos ( π n + 1 ) ) = ( cos ( π 3 ) − cos ( π 4 ) ) + ( cos ( π 4 ) − cos ( π 5 ) ) + ⋯ + ( cos ( π N ) − cos ( π N + 1 ) ) = cos ( π 3 ) − cos ( π N + 1 ) . {\displaystyle {\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 → ∞ {\displaystyle N\to \infty } , the argument π N + 1 {\displaystyle {\frac {\pi }{N+1}}} converges to 0 {\displaystyle 0} , and cos x {\displaystyle \cos x} is continuous at x = 0 {\displaystyle x=0} .
Thus the value of the series is lim N → ∞ s N = lim N → ∞ ( cos ( π 3 ) − cos ( π N + 1 ) ) = cos ( π 3 ) − cos ( lim N → ∞ π N + 1 ) = cos ( π 3 ) − cos ( 0 ) = − 1 2 . {\displaystyle \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 − 1 2 . {\displaystyle -{\frac {1}{2}}.}
