The limit of a sum

Question

I'm trying to find out this limit $$\lim_{n \rightarrow \infty} \frac{1}{n}\sum_{k=1}^n e^{-\Large\frac{k\pi\sqrt{2}}{4n}}\left(\tan\frac{k\pi}{4n}\right)^2 =?$$ My try: I know I have to transform the sum into a Riemann sum, but I've found nothing useful. Any help would be great.

Answer 1

Hint: You get, using a Riemann sum, as $n \longrightarrow +\infty$, $$\frac{1}{n}\sum_{k=1}^n e^{-\Large\frac{\sqrt{2}\:k\pi}{4n}}\left(\tan\frac{k\pi}{4n}\right)^2 \longrightarrow \frac{4}{\pi}\int_0^{\Large\frac{\pi}{4}}e^{-\sqrt{2}\:x}\left(\tan x\right)^2 dx$$