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Question Number 57405 by Abdo msup. last updated on 03/Apr/19

$$calculate{lim}_{x\to 0}\frac{1-cosx.cos\left(2x\right)....cos\left(nx\right)}{x^2}$$$$withnintegrnaturalnot0.$$

Answered by Smail last updated on 05/Apr/19

$$cosx=1-\frac{x^2}{2}+o\left(x^2\right)$$$$cos2x=1-\frac{{\left(2x\right)}^2}{2}+o\left(x^2\right)$$$$cos3x=1-\frac{{\left(3x\right)}^2}{2}+o\left(x^2\right)$$$$cosx.cos\left(2x\right).cos\left(3x\right)...cos\left(nx\right)=(1-\frac{x^2}{2})(1-\frac{{\left(2x\right)}^2}{2})...(1-\frac{{\left(nx\right)}^2}{2})+o\left(x^2\right)$$$$=1-(\frac{x^2}{2}+\frac{{\left(2x\right)}^2}{2}+\frac{{\left(3x\right)}^2}{2}+...\frac{{\left(nx\right)}^2}{2})+o\left(x^2\right)$$$$=1-\frac{x^2}{2}(1+2^2+3^2+...+n^2)+o\left(x^2\right)$$$$1-cosx.cos2x...cos\left(nx\right)\underset{0}{\sim }\frac{x^2}{2}\left(\frac{n(n+1)(2n+1)}{6}\right)$$$$\frac{1-cosx.cos\left(2x\right)...cos\left(nx\right)}{x^2}\underset{0}{\sim }\frac{n(n+1)(2n+1)}{12}$$$$\underset{x\to 0}{lim}\frac{1-cosx.cos\left(2x\right)...cos\left(nx\right)}{x^2}=\frac{n(n+1)(2n+1)}{12}$$

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