I am trying to prove that the limit superior of $\cos(x^2)-\cos[(x+1)^2]=2$ as $x \rightarrow \infty.$ I was trying to construct a subsequence that converges to 2 (along the lines of $\sqrt{2 \pi k}$) to no avail. Appreciate if anyone can offer any hints on how to proceed with this.


Here is the general fact: if $a/b$ is irrational and $c$ is an arbitrary real number, then in the sequence $c+na,n\in\mathbb N$ there are elements close enough to a multiple of $b$ for arbitrary large $n$. To prove this, first note that we can divide $a$ and $c$ by $b$ and so assume $b=1$, $a$ irrational. Then consider the sequence $\{c+na\}$ - the fractional part of $c+na$. Because of irrationality of $a$ no two elements of this sequence coincide, but if we subdivide the segment $[0,1]$ into small pieces of length $\varepsilon$, there are, of course, those $c+na$ and $c+ma$ whose fractional parts lie in the same piece, so $(n-m)a$ is $\varepsilon$-close to an integer, but not an integer. Then we can find $k$ such that $c+k(n-m)a$ is $\varepsilon$-close to an integer.

Now we need in fact $\pi+(\sqrt{2\pi k}+1)^2$ to be arbitrarily close to some multiple of $2\pi$ (then the second $\cos$ is close to $-1$). But this is equivalent for $(1+\pi)+2\sqrt{2\pi k}$ to be close to a multiple of $2\pi$. If we take $k=n^2,c=1+\pi,a=2\sqrt{2\pi},b=2\pi$ this will follow from the fact above and irrationality of $\pi$ which is not easy to prove, but at least well-known.

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  • $\begingroup$ Much appreciated! $\endgroup$ – daniel Jan 29 '17 at 17:12

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