Sequence $\sin^2n$ I would like to prove that if $a_n = \sin^2(n),$ then it does not converge. 
Usually we show 2 subsequences with different limits for those sine cases, but I could not do this since my n is a natural number and I cant construct those subsequences. How can I show, then, it doesn't have limit?
Thanks in advance!
 A: There is always an integer $a_k$ between $2k\pi + \frac{\pi}{4}$ and $2k\pi + \frac{3\pi}{4}$
Consider that sequence. $\sin^2 a_k > \frac{1}{2}$ for all $k$.
There is always an integer $b_k$ between $2k\pi + \frac{5\pi}{6}$ and $2k\pi + \frac{7\pi}{6}$
Consider that sequence. $\sin^2 b_k < \frac{1}{4}$ for all $k$.
A: One way to prove it is to show that the sequence is not Cauchy. If it were Cauchy, you'd have 
$|a_{n+1}-a_n|< \epsilon$ for any $\epsilon > 0$ as long as $n$ is large enough. To show that this is not the case you can see that
$|a_{n+1}-a_n| = |\textrm{sin }(1+n)^2 - \textrm{sin }(n)^2|$ is not a sequence that goes to zero (Use the addition formula for sine). 
http://mathworld.wolfram.com/TrigonometricAdditionFormulas.html
A: Assume instead that $\displaystyle\lim_{n\to\infty}\sin^{2}n=L$.  Then $\displaystyle\lim_{n\to\infty}\cos^{2}n=1-L$, and $\displaystyle\lim_{n\to\infty}\sin^{2}n=\lim_{n\to\infty}\frac{1}{2}(1-\cos 2n)=L\implies\lim_{n\to\infty}\cos 2n=1-2L$; so $\displaystyle\lim_{n\to\infty}\cos^{2}2n=1-L\implies(1-2L)^2=1-L\implies4L^2-3L=0\implies L=0 \text{ or } L=\frac{3}{4}$.
$\textbf{1)}$ If $L=0$, then $\displaystyle\lim_{n\to\infty}\sin n=0$ and $\displaystyle\lim_{n\to\infty}\cos^{2}n=1$ and $\displaystyle\lim_{n\to\infty}\cos 2n=\lim_{n\to\infty}(\cos^{2}n-\sin^{2}n)=1\implies$
$\lim_{n\to\infty}\cos(2n+1)=\lim_{n\to\infty}\left(\cos 2n\cos1-\sin 2n\sin1\right)=\cos1\implies$
$\lim_{n\to\infty}\cos^{2}(2n+1)=(\cos1)^2=1,$ which gives a contradiction.
$\textbf{2)}$  If $L=\frac{3}{4}$, then $\displaystyle\lim_{n\to\infty}\sin^{2}n=\frac{3}{4}$ and $\displaystyle\lim_{n\to\infty}\cos^{2}n=\frac{1}{4}$, so
$\displaystyle\lim_{n\to\infty}(\cos(n+1))^2=\lim_{n\to\infty}(\cos n\cos1-\sin n\sin1)^2$
$\displaystyle=\lim_{n\to\infty}(\cos^{2}n\cos^{2}1-\sin 2n\sin1\cos1+\sin^{2}n\sin^{2}1)$
$\displaystyle=\frac{1}{4}\cos^{2}1-(\sin1\cos1)[\lim_{n\to\infty}\sin 2n]+\frac{3}{4}\sin^{2}1=\frac{1}{4},$ 
so
$\displaystyle(4\sin1\cos1)[\lim_{n\to\infty}\sin 2n]=\cos^{2}1+3\sin^{2}1-1=2\sin^{2}1\implies\lim_{n\to\infty}\sin2n=\frac{1}{2}\tan1\implies\lim_{n\to\infty}\sin^{2}2n=\frac{1}{4}\tan^{2}1=\frac{3}{4}\implies\tan^{2}1=3,\text  {which gives a contradiction.}$
A: It is known that if you step along a circle, with an irrational ratio between the length of your step and the perimeter of the circle, then the set of all points along your way is dense in the perimeter. In particular, if the length of your step is $1$, your going to be very close to both $\pi/2$ and $0$ infinitely many times. This is enough for the above series not to converge.
A: Here's a proof without finding subsequences. Suppose that $\sin^2 n\to a$, $n\to\infty$. Then $\cos^2n\to1-a$. Since $\sin^22n=4\sin^2n\cos^2n$, taking limit we obtain $a=4a(1-a)$, so either $a=\frac34$ or $a=0$. In the first case $$\sin^23n=\sin^2n(3-4\sin^2n)^2\to 0,$$ a contradiction. In the second case $\sin n\to 0$ and $$\sin(n+1)=\sin n\cos 1+\cos n\sin 1\to 0,$$
so also $\cos n\to 0$, which obviously contradicts the basic trigonometric identity.
