주어진 급수를 변형하면
lim n → ∞ π n ∑ k = 1 n cos k π 4 n = 4 lim n → ∞ π 4 n ∑ k = 1 n cos k π 4 n \displaystyle \lim\limits _{n \rightarrow \infty} \frac{\pi}{n} \sum\limits _{k = 1} ^{n} \cos \frac{k \pi}{4 n} = 4 \lim\limits _{n \rightarrow \infty} \frac{\pi}{4 n} \sum\limits _{k = 1} ^{n} \cos \frac{k \pi}{4 n} n → ∞ lim n π k = 1 ∑ n cos 4 n k π = 4 n → ∞ lim 4 n π k = 1 ∑ n cos 4 n k π
f ( x ) = cos x f \left( x \right) = \cos x f ( x ) = cos x , x k = k π 4 n \displaystyle x _{k} = \frac{k \pi}{4 n} x k = 4 n k π , δ x = π 4 n \displaystyle \delta x = \frac{\pi}{4 n} δ x = 4 n π 라 하면 정적분과 급수의 관계에 의하여
4 lim n → ∞ π 4 n ∑ k = 1 n cos k π 4 n \displaystyle 4 \lim\limits _{n \rightarrow \infty} \frac{\pi}{4 n} \sum\limits _{k = 1} ^{n} \cos \frac{k \pi}{4 n} 4 n → ∞ lim 4 n π k = 1 ∑ n cos 4 n k π = 4 lim n → ∞ ∑ k = 1 n f ( x k ) Δ x \displaystyle = 4 \lim\limits _{n \rightarrow \infty} {} \sum\limits _{k = 1} ^{n} f \left( x _{k} \right) \Delta x = 4 n → ∞ lim k = 1 ∑ n f ( x k ) Δ x
= 4 ∫ 0 π 4 cos x d x \displaystyle = 4 \int _{0} ^{\frac{\pi}{4}} \cos x dx = 4 ∫ 0 4 π cos x d x
= 4 [ sin x ] 0 π 4 \displaystyle = 4 \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \sin x \end{array} \right] _{0} ^{\frac{\pi}{4}} = 4 [ sin x ] 0 4 π
= 4 ( sin π 4 − sin 0 ) \displaystyle = 4 \left( \sin \frac{\pi}{4} - \sin 0 \right) = 4 ( sin 4 π − sin 0 )
= 4 × 2 2 \displaystyle = 4 \times \frac{\sqrt{2}}{2} = 4 × 2 2 = 2 2 \displaystyle = 2 \sqrt{2} = 2 2