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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.

claudiubagiu
  • 97
  • use that $\sum^n_{k=1}\cos\frac{2k}{n\sqrt{n}}$ is more or less a geometric series (the real part of one); or compute directly from trigonometric identities $S=\sum^n_{k=1}\cos\frac{2k}{a}$ by multiplying both sides with $\cos\frac{2}{a}$ and plug in $a=\frac{1}{n\sqrt{n}}$ to get a closed form for the sum; then apply usual manipulations – Conrad May 01 '22 at 16:13
  • You may wish to use write $n-\sum_{k=1}^n\cos\left(\frac{2k}{n\sqrt{n}}\right)$ as $\sum_{k=1}^n\left(1-\cos\left(\frac{2k}{n\sqrt{n}}\right)\right)$ and then use the fact that $\Big|1-\cos(x)-\frac{x^2}{2!}\Big|\leq \frac{x^4}{4!}$ for all $x$. –  May 01 '22 at 16:16
  • $1-\cos(2x)=2\sin^2x,$ $x={k\over n\sqrt{n}}.$ – Ryszard Szwarc May 01 '22 at 16:21

2 Answers2

6

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.

Yuval Peres
  • 22,782
1

You face the sum of cosines with angles in arithmetic progression.

So $$\sum_{k=1}^n \cos \left(\frac{2 k}{n^{3/2}}\right)=\sin \left(\frac{1}{\sqrt{n}}\right) \cos \left(\frac{n+1}{n^{3/2}}\right) \csc \left(\frac{1}{n^{3/2}}\right)$$

$$S_n=n-\sin \left(\frac{1}{\sqrt{n}}\right) \cos \left(\frac{n+1}{n^{3/2}}\right) \csc\left(\frac{1}{n^{3/2}}\right)$$ Now, using Taylor series $$S_n=\frac{2}{3}+\frac{13}{15 n}+\frac{4}{315 n^2}+O\left(\frac{1}{n^3}\right)$$

If $n>6$, the relative error is smaller than $0.1$%. If $n>14$, it becomes smaller than $0.01$%.

Claude Leibovici
  • 289,558