Show that $\frac{\cos^2(n)}{1+n^2}$ is not decreasing. Let $a_n=\frac{\cos^2(n)}{1+n^2}$.  I want to show that the sequence $(a_n)_n$ never becomes decreasing, i.e, if $b^k_n=a_{n+k}$, then, the sequence $(b^k_n)_n$ is not decreasing for all $k\in\mathbb{N}$. Although we have the cosine function oscilating there, it is easy to see that we can construct decreasing sequences out of $f(x)=\frac{\cos^2(x)}{1+x^2}$. It seems like the sets $A=\{x\in\mathbb{R}\,|\,f(x)>f(x+1)\}$ and $A^c$ both have infinite intersection with $\mathbb{N}$, and that would be sufficient to show what i want, but i don't know how to prove that. any tips will be aprecciated.
 A: Find an $n$ such that $n$ is within $\varepsilon$ of $\frac{\pi}{2}+k\pi$ for some integer $k$, where $\varepsilon$ is, say, $0.01$. You can do this because when you reduce integers mod $2\pi$, an irrational number, into $[0,2\pi)$, the set is dense.
This tells you $\cos(n)$ is within $\varepsilon$ of $0$. Then $a_n=\frac{\cos^2(n)}{1+n^2}<\frac{\varepsilon^2}{1+n^2}$.
Meanwhile $n+1$ is within $\varepsilon$ of $\frac{\pi}{2}+1+k\pi$ for some integer $k$. This tells you $\left\lvert\cos(n+1)\right\rvert$ is roughly $\left\lvert\cos(\pi/2+1)\right\rvert$, but certainly greater than $\cos(1)$.

So $a_{n+1}>\frac{\cos(1)^2}{1+(n+1)^2}$.
Now is $\frac{\cos(1)^2}{1+(n+1)^2}>\frac{\varepsilon^2}{1+n^2}$? That is equivalent to $$\left(\cos^2(1)-\varepsilon^2\right)n^2-\varepsilon^2n +\cos^2(1)-2\varepsilon^2>0$$
For a fixed small $\varepsilon$, this is quadratic in $n$. So there are infinitely many $n$ that satisfy this inequality and are also within $\varepsilon$ of $\frac{\pi}{2}+k\pi$.
So we can always find $n$ where $a_{n+1}>a_n$.
A: In order to show that
$${\cos^2n\over1+n^2}-{\cos^2(n+1)\over1+(n+1)^2}$$
is infinitely often negative, it suffices to show the same for
$$(1+n^2)(\cos^2n-\cos^2(n+1))+(2n+1)\cos^2n$$
Note that
$$\begin{align}
\cos^2n-\cos^2(n+1)
&=\cos^2n-(\cos n\cos1-\sin n\sin1)^2\\
&=\cos^2n(1-\cos^21)+2\cos n\sin n\cos1\sin1-\sin^2n\sin^21\\
&=\cos^2n\sin^21+\sin2n\cos1\sin1-\sin^2n\sin^21\\
&=(\cos2n\sin1+\sin2n\cos1)\sin1\\
&=\sin(2n+1)\sin1
\end{align}$$
So we need to show that
$$(1+n^2)\sin(2n+1)\sin1+(2n+1)\cos^2n$$
is infinitely often negative.
Since $2\pi/3\gt2$, it happens that $2n+1\in(7\pi/6,11\pi/6)$ mod $2\pi$ for infinitely many $n$, and for each such $n$, we have  $\sin(2n+1)\lt-1/2$. And since $\sin1\gt0$, we have $(n^2+1)\sin1\gt2(2n+1)$ for all sufficiently large $n$. It follows that
$$(1+n^2)\sin(2n+1)\sin1+(2n+1)\cos^2n$$
is negative for infinitely many $n$, and we're done.
(My thanks to the OP for pointing out a mistake in the original answer.)
A: Define
$$f(x):=\frac{\cos^2x}{1+x^2}\implies f'(x)=\frac{-\sin2x\cdot(1+x^2)-2x\cos^2x}{(1+x^2)^2}¿$$
$$=-\frac{(1+x^2)\sin2x+2x(\cos 2x+1)}{(1+x^2)^2}$$
Now, the above is negative (and thus $\;f\;$ is monotone descending) iff
$$(1+x^2)\sin2x+2x\cos 2x +2x>0\;,\;\;x>0$$
and it's easy to see this is false, for example: for $\;x_0=4,000,003\cdot\cfrac\pi4\;$, we get
$$(1+x_0^2)\cdot(-1)+0+2x_0=-x_0^2+2x_0-1=-(x_0-1)^2<0$$
and of course the above happens infinite times...
