How to find the limit of $\cos^2(\pi\sqrt{n^2-n})$?

Question

I have been ask to find the limit $\lim \limits_{n \to \infty} \cos^2(\pi\sqrt{n^2-n})$

However surely as $n$ gets larger the expression inside the $\cos$ function tends to infinity and the whole thing oscillates between $0$ and $1$?

score 1 · Answer 1 · answered Sep 14 '15 at 17:48

1

Hint:

Use the fact that for all integers $n$ we have $\cos^2(x)=\cos^2(x-\pi n)$.

answered Sep 14 '15 at 17:48

Jyrki Lahtonen

133,153

Fairly sure that this is essentially a duplicate. Gimme a minute. – Jyrki Lahtonen Sep 14 '15 at 17:48
1

http://math.stackexchange.com/questions/46065/lim-n-to-infty-cos-pi-sqrtn2-n-second-battle – Anthony Peter Sep 14 '15 at 17:49
Good job @AnthonyPeter. The key idea is explained there. – Jyrki Lahtonen Sep 14 '15 at 17:56

score 1 · Answer 2 · answered Sep 14 '15 at 17:50

1

A start: First show that $\lim_{n\to\infty}(n-\sqrt{n^2-n})=\frac{1}{2}$. So for large $n$, the cosine of $\pi\sqrt{n^2-n}$ is close to the cosine of $\pi n-\frac{\pi}{2}$.

answered Sep 14 '15 at 17:50

André Nicolas

507,029

How to find the limit of $\cos^2(\pi\sqrt{n^2-n})$?

2 Answers2