Inequality and sequence of functions I am trying to prove that the sequence of functions $f_n(x)=\sqrt{n}\cos(nx)-\frac{\sin(nx)}{2n\sqrt{n}}$ does not converge for any $x\in\mathbb{R}$.
My idea was to fix $x\in\mathbb{R}$ and then obtain a bound like $|f_n(x)|\geq $ constant $\times$ (function of $n$ which goes to infinity as $n\to\infty$) so $|f_n(x)|=|\sqrt{n}\cos(nx)-\frac{\sin(nx)}{2n\sqrt{n}}|\geq ||\sqrt{n}\cos(nx)|-|\frac{\sin(nx)}{2n\sqrt{n}}||\geq \sqrt{n}||\cos(nx)|-\frac{1}{2n^2}|$ which is almost what I wanted but I don't see how to go further so I would appreciate an hint. Thank you.
As the sequence of functions $\{g_n\}$ with
$$g_n(x)=\frac{\sin(nx)}{2n\sqrt{n}}$$ converges to $0$ for all $x \in \mathbb R$, it is enough to prove that $\{h_n\}$ does not converge for any $x \in \mathbb R$ with
$$h_n(x) = \sqrt{n}\cos(nx).$$
And this is clear as $\cos nx$ never converges to zero which would be required for $\{h_n(x)\}$ to converge as $\lim\limits_{n \to \infty} \sqrt n = \infty$.
Proof that $\{\cos nx\}$ never converges to zero:
As $$\cos(n+1)x = \cos nx \cos x - \sin nx \sin x,$$ if $\{ \cos nx \}$was converging to zero, we would have
$$\lim\limits_{n \to \infty} \sin x \sin nx = 0$$
which implies either $\sin x = 0$, i.e. $x \in {\pi \mathbb Z}$ and a contradiction as $\{\cos np\pi\}$ doesn't converge to zero. Or $\lim\limits_{n \to \infty} \sin nx = 0$ and another contradiction as
$$\cos^2 nx + \sin^2 nx = 1.$$
$f_n(x)=\sqrt{n}\cos(nx)-\frac{\sin(nx)}{2n\sqrt{n}}$
$\frac{\sin(nx)}{2n\sqrt{n}} \rightarrow 0$ when $n\rightarrow +\infty$.
And $\sqrt{n} \cos (nx)$ diverges (we can easily extract a subsequence that diverges to $+\infty$)