lim-n-k-n-2n-sin-pi-k-

Question Number 166699 by qaz last updated on 25/Feb/22

\lim_{n \to \infty} \underset{k = n}{\sum^{2 n}} \sin \frac{π}{k} = ?

Answered by mathsmine last updated on 25/Feb/22

x - \frac{x^{3}}{6} ⩽ s i n (x) ⩽ x \dots \dots E

\underset{k = n}{\sum^{2 n}} \frac{π}{k} = \underset{k = 0}{\sum^{n}} \frac{π}{n + k} = \frac{1}{n} \underset{k = 0}{\sum^{n}} \frac{π}{1 + \frac{k}{n}} = S_{n}

\lim_{n \to \infty} S_{n} = π \int_{0}^{1} \frac{d x}{1 + x} = π l n (2)

\underset{k = n}{\sum^{2 n}} {(_{} \frac{π}{n + k})}^{3} = T_{n} ⩽ (n + 1) {(\frac{π}{2 n})}^{3} = \frac{π^{3}}{8} (\frac{1}{n^{2}} + \frac{1}{n^{3}})

E \Rightarrow

S_{n} - \frac{T_{n}}{6} ⩽ \underset{k = n}{\sum^{2 n}} s i n (\frac{π}{k}) ⩽ S_{n}

\lim_{n \to \infty} S_{n} - \frac{T_{n}}{6} ⩽ \lim_{n \to \infty} \underset{n}{\sum^{2 n}} s i n (\frac{π}{k}) ⩽ \lim_{n \to \infty} S_{n}

\lim_{n \to \infty} \underset{k = n}{\sum^{2 n}} s i n (\frac{π}{k}) = π l n (2)