How to calculate $\lim\limits_{n\rightarrow \infty }(n-\sum^n_{k=1}\cos\frac{2k}{n\sqrt{n}}) $? I tried this:
$\lim\limits_{n\rightarrow \infty }(n-\sum^n_{k=1}\cos\frac{2k}{n\sqrt{n}}) $=
= $ \lim\limits_{n\rightarrow \infty }(1-\cos\frac{2}{n\sqrt{n}}) $ + $ \lim\limits_{n\rightarrow \infty }(1-\cos\frac{4}{n\sqrt{n}}) $ + ... + $ \lim\limits_{n\rightarrow \infty }(1-\cos\frac{2n}{n\sqrt{n}}) $ =
= $ \lim\limits_{n\rightarrow \infty }\frac{(1-\cos\frac{2}{n\sqrt{n}})(1+\cos\frac{2}{n\sqrt{n}})}{1+\cos\frac{2}{n\sqrt{n}}} $ + $ \lim\limits_{n\rightarrow \infty }\frac{(1-\cos\frac{4}{n\sqrt{n}})(1+\cos\frac{4}{n\sqrt{n}})}{1+\cos\frac{4}{n\sqrt{n}}} $ + ...+ $ \lim\limits_{n\rightarrow \infty }\frac{(1-\cos\frac{2n}{n\sqrt{n}})(1+\cos\frac{2n}{n\sqrt{n}})}{1+\cos\frac{2n}{n\sqrt{n}}} $ =
= $ \lim\limits_{n\rightarrow \infty }\frac{\sin^2\frac{2}{n\sqrt{n}}}{1+\cos\frac{2}{n\sqrt{n}}} $ + $ \lim\limits_{n\rightarrow \infty }\frac{\sin^2\frac{4}{n\sqrt{n}}}{1+\cos\frac{4}{n\sqrt{n}}} $ + ... + $ \lim\limits_{n\rightarrow \infty }\frac{\sin^2\frac{2n}{n\sqrt{n}}}{1+\cos\frac{2n}{n\sqrt{n}}} $ = ...
, but from here I don't know what to do.

Call the required limit $L$. Use the Taylor expansion
$$\cos t=1-t^2/2+O(t^4)$$ to deduce that
$$L=\lim_n \sum_{k=1}^n [2(k/n)^2 n^{-1}+O(n^{-2})] =\int_0^1 2x^2 \,dx=2/3 \,,$$
since the sum is a Riemann sum.
