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Calculus Level 3

lim n 1 n k = 1 n cos ( k π 2 n ) \large \lim_{n\to\infty} \frac1n \sum_{k=1}^n \cos\left(\frac{k\pi}{2n}\right)

If the limit above equals to A π B A \pi^B for integers A A and B B , find the value of A × B A\times B .


The answer is -2.

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1 solution

The given limit of summation is a Riemann Integral.

I = 0 1 cos ( π 2 x ) d x I = \displaystyle \int_{0}^{1} \cos\left(\dfrac{\pi}{2} x\right) dx

t = π 2 x d t = π 2 d x t = \dfrac{\pi}{2}x \Rightarrow dt = \dfrac{\pi}{2} dx

I = 2 π 0 π 2 cos t d t = 2 π I = \displaystyle \dfrac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \cos t dt = \dfrac{2}{\pi}

A = 2 , B = 1 A × B = 2 A = 2 , B = -1 \Rightarrow A\times B = \boxed{-2}

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