A more succinct way of showing that $\{f_n(x)\}_n$ does not converge for any $x \in [0,1].$

Question

Here is the proof I have:

Is there a more succinct (elegant and organized) way of showing that $\{f_n(x)\}_n$ does not converge for any $x \in [0,1]$?

Could someone help me with this, please?

Answer 1

Fix an $x\in(0,1]$, for each $m\in \Bbb N$ choose the smallest $n$ such that

$$nx+1-m\pi=a≥0.$$

Since $x\in(0,1]$, it is obvious that $a<1$. Using the fact $\sin(x-\pi)=-\sin x$, we have

$$|\sin (nx+1)|=|\sin a|<\sin 1.$$

Thus there are infinitely many $n$ such that $|f_n(x)|<(x+1)\sin 1$ and therefore if the sequence converges, it would converge to some number with absolute value at most $(x+1)\sin 1.$

But applying the same strategy also shows that there are also infinitely many $n$ such that $|f_n(x)|>(x+1)\sin (\pi/2-1)$ consequently, the sequence does not converge.