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Question Number 88065 by arcana last updated on 08/Apr/20

$$\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{i=1}{\sum ^n}\cos {}^2\left(\frac{\pi i}{n}\right)$$

Commented by mathmax by abdo last updated on 08/Apr/20

$${\int }_a^{b}f\left(x\right)dx={lim}_{n\to +\mathrm{\infty }}\frac{b-a}{n}\sum _{k=1}^{n}f(a+\frac{k(b-a)}{n})\left(Riemansum\right)\Rightarrow$$$${lim}_{n\to \mathrm{\infty }}\frac{1}{n}\sum _{i=1}^n{cos}^2\left(\frac{i\pi }{n}\right)=\frac{1}{\pi }{lim}_{n\to +\mathrm{\infty }}\frac{\pi -0}{n}\sum _{i=1}^n{cos}^2\left(\frac{i(\pi -0)}{n}\right)$$$$=\frac{1}{\pi }{\int }_0^{\pi }{cos}^{2}xdx=\frac{1}{2\pi }{\int }_0^{\pi }(1+cos\left(2x\right))dx$$$$=\frac{1}{2}+\frac{1}{4\pi }{\left[sin\left(2x\right)\right]}_0^{\pi }=\frac{1}{2}$$

Commented by arcana last updated on 08/Apr/20

$$gracias$$

Commented by abdomathmax last updated on 10/Apr/20

$$youarewelcome$$

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