Calculus problem by Sutirtha Datta

$\begin{aligned} L & = \lim_{n \to \infty} \frac 1n \sum_{k=0}^{n-1} \cos \left( \frac {k \pi}{2n} \right) & \small \color{#3D99F6} \text{By Riemann's sum: } \lim_{n \to \infty} \frac 1n \sum_{k=1}^n f \left( \frac kn \right) = \int_a^b f(x) \ dx \\ & = \int_0^1 \cos \left( \frac {\pi x}2 \right) \ dx & \small \color{#3D99F6} a = \lim_{n \to \infty} \frac 0n = 0, \ b = \lim_{n \to \infty} \frac {n-1}n = 1 \\ & = \frac 2 \pi \sin \left( \frac {\pi x}2 \right) \bigg|_0^1 \\ & = \frac 2 \pi \end{aligned}$

$\implies a + b = 2 + 1 = \boxed{3}$

$\lim_{n\rightarrow \infty}\frac{1}{n} \sum_{k=0}^{n-1}\cos(\frac{k\pi}{2n})$

= ${\frac{2}{\pi}}\lim_{n\rightarrow \infty}\frac{1}{\frac{2n}{\pi}} \sum_{k=0}^{n-1}\cos(\frac{k\pi}{2n})$

= ${\frac{2}{\pi}}\int_a^b{\cos{x}{dx}}$

[Where $a= \lim_{n\rightarrow \infty}\frac{0}{\frac{2n}{\pi}}$ , $b= \lim_{n\rightarrow \infty}\frac{(n-1)}{\frac{2n}{\pi}}$ , $x=\frac{k}{\frac{2n}{\pi}}$ , $dx=\frac{1}{\frac{2n}{\pi}}$ ]

= $\frac{2}{\pi}\sin{\frac{\pi}{2}}$ = $\frac{2}{\pi}$