# Finding $\lim_{n\rightarrow\infty} \sum^{n}_{k=0}\left|\frac{2\pi\cos(k\pi(3-\sqrt{5}))}{n}\right|$

Finding $\displaystyle \lim_{n\rightarrow\infty} \sum^{n}_{k=0}\left|\frac{2\pi\cos(k\pi(3-\sqrt{5}))}{n}\right|$

Try: Assuming $$S=\lim_{n \rightarrow\infty}\frac{2\pi}{n}\sum^{n}_{k=0}\left|\cos(k\pi(3-\sqrt{5})\right)|$$=

$$\lim_{n\rightarrow \infty}\frac{2\pi}{n}\sum^{n}_{k=0}\Re{e^{i\left(k\pi(3-\sqrt{5})\right)}}$$

Could some help me to solve it, Thanks

• Do you need to prove the limit is zero? – DonAntonio Jan 17 '18 at 0:21
• @Durgesh Tiwari: Why did you drop the absolute value in your last step? – jgsmath Jan 17 '18 at 0:34
• Also, if you don't want to consider absolute value, you could write $\cos{x} = (e^{ix}+e^{-ix})/2$. Then the summation becomes a simple addition of two finite geometric series with their corresponding common ratio. – jgsmath Jan 17 '18 at 0:38

Since $\pi(3-\sqrt{5})$ is irrational, the equidistribution theorem gives

$$\lim_{n \rightarrow\infty}\frac{1}{n+1}\sum^{n}_{k=0}\left|\cos(k\pi(3-\sqrt{5})\right)| = \int_0^1|\cos \pi x|\, dx = \frac{2}{\pi}.$$

It follows that the limit in this problem equals $2\pi \dfrac{2}{\pi} =4.$

What you mind find useful is the following Theorem:

$(a_n)_{n\in\mathbb{N}}$ $\rightarrow$ A as $n\rightarrow\infty$ then $\frac{1}{n} \sum_{k=0}^na_k \rightarrow A$ as $n\rightarrow\infty$. (Cauchycher Grenzwertsatz) .

The follwing holds also:

$lim_{n\to\infty}|a_n|=|lim_{n\to\infty}a_n|$.

At the end the sum shall converge to $2\pi$.

• 21saw plese explain me in detail. – DXT Jan 17 '18 at 3:40
• How could that be true? You have a bunch of numbers in between $0$ and $2\pi.$ It would be unusual if their limiting average value is $2\pi.$ – zhw. Jan 17 '18 at 5:20
• you are indeed right. The problem is that $|(cos(k\pi(3-5^.5)))|$ doesn't even converge as k approaches $\infty$. Sorry for causing even more confusion – 21säv Jan 17 '18 at 9:30