I'd love your help with finding the following limit: $$\lim_{n\to \infty }\cos (\pi\sqrt{n^{2}-n}).$$

I was asked to find this limit, but honestly I believe that it doesn't exist.

According to Heine Theorem of limit of functions, I can choose two sequences:

$x_{k}=2\pi k$ and $y_{k}=2\pi k+\pi$ and notice that when I apply the function on both of them, I'll get -1 and 1, respectively.

Am I right?

Thank you again.

  • $\begingroup$ Do you mean to have that $\pi$ in your sequences? There's already multiplication by $\pi$ inside the cosine. Or did you plan to start with, say, $x_k$, and work backwards to a sequence of $n_k$ such that $\cos\pi\sqrt{n_k^2-n_k} = \cos{x_k}$? $\endgroup$ Commented Jun 16, 2011 at 17:12
  • $\begingroup$ Yes, my mistake. without the $\pi$'s. thanks. $\endgroup$
    – user6163
    Commented Jun 17, 2011 at 7:05
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    $\begingroup$ You kinda started the wrong way, by asking what tool to use. Your first question should have been "what's happening here?" It is then natural to calculate, for largish but not too large $n$. (Not too large because if you take $n=10^9$, roundoff error kills you.) So look at $n=100, 101, 102$. Calculate. You will get answers near $0$. And while you are taking $\sqrt{n^2-n}$, you may notice it is almost exactly halfway between consecutive integers. Now an idea for a proof may come. $\endgroup$ Commented Jun 18, 2011 at 3:39

3 Answers 3


Since a nice formal argument has been supplied by Davide Giraudo, I will allow myself the luxury of informality.

Let $n$ be a large positive integer. Complete the square. We have $$n^2-n=\left(n-\frac{1}{2}\right)^2 -\frac{1}{4}$$

Take the square root. When $n$ is very large, the term $-1/4$ makes a vanishingly small contribution to the square root.

So our square root is nearly equal to $n-1/2$. And the cosine of $\pi n -\pi/2$ is $0$.


We have \begin{align*} \cos (\pi\sqrt{n^2-n})&= (-1)^n\cos(\pi(\sqrt{n^2-n}-n))\\ &= (-1)^n \cos\pi\frac{-n}{\sqrt{n^2-n}+n}\\ &=(-1)^n\cos \pi\frac 1{\sqrt{1-\frac 1n }+1}, \end{align*} hence $|\cos(\pi\sqrt{n^2-n})| = \left|\cos \left(\pi\frac 1{\sqrt{1-\frac 1n }+1}\right)\right|$. Since $\lim \limits_{n\to +\infty}\pi\frac 1{\sqrt{1-\frac 1n }+1} =\frac{\pi}2$, the $\cos$ is continuous and $\cos \frac{\pi}2 =0$ we conclude that the limit is $0$.

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    $\begingroup$ It is intresting that when you go by intuition and write the limit as $\cos \pi n\sqrt{1-\frac{1}{n}}$ you sincerely might think that the limit does not exist. $\endgroup$
    – user6163
    Commented Jun 18, 2011 at 10:14
  • $\begingroup$ @Nir Yes, I agree. The idea comes from an exercise that I've done some years ago. I had to look at the convergence of the series $\sum_{n\geq 0}\sin(\pi\sqrt{n^2+1})$. $\endgroup$ Commented Jun 18, 2011 at 10:25
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    $\begingroup$ Are you sure of this answer? wolframalpha.com/input/?i=limit+cos%28pi*sqrt+%28n^2-n%29%29 $\endgroup$
    – user6163
    Commented Jun 18, 2011 at 10:37
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    $\begingroup$ @Nir: it is very important that $n$ here is integer. $\endgroup$
    – Marek
    Commented Jun 18, 2011 at 12:18

Considering the form $\cos(\pi n \sqrt{1-\frac{1}{n}})$ and using Taylor's expansion for $\sqrt{1-\frac{1}{x}}$ $\to$ see here, we get that when n is large $\cos (\pi\sqrt{n^{2}-n}) \approx \cos( \pi n -\frac{\pi}{2})$. Therefore, $L=0$.


  • $\begingroup$ Nice answer Chris's sis........ $\endgroup$
    – juantheron
    Commented Nov 3, 2013 at 13:36

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