Determine if the sequence functions $f_n(x)=\frac{\sin(nx)+\cos(nx)}{\ln(n)}$ uniformly converges by Cauchy's criterion in $x\in[e,\infty)$ Determine if the sequence functions $f_n(x)=\frac{\sin(nx)+\cos(nx)}{\ln(n)}$ uniformly converges by Cauchy's criterion in $x\in[e,\infty)$
Attempt:
Let $0<\varepsilon $ we chose $\displaystyle n^{*} =\left\lceil \frac{1}{\varepsilon }\right\rceil +1$
Therefore, for all $n^*<n,k$ and for all $x\in[e,\infty)$ exists
\begin{aligned}
\Bigl|\frac{\sin( nx) +\cos( nx)}{\ln( n)} -\frac{\sin( kx) +\cos( kx)}{\ln( k)}\Bigl| & \leq \Bigl|\frac{\sin( nx) +\cos( nx)}{\ln( n)}\Bigl|\\
 & \leq \Bigl|\frac{\sin^{2}( nx) +\cos^{2}( nx)}{\ln( n)}\Bigl|\\
 & \leq \Bigl|\frac{1}{\ln( n)}\Bigl|\\
 &  < \varepsilon 
\end{aligned}
I noticed I have a few mistakes with my attempt and I'm unable to think of another direction to solve this question by Cauchy's Criterion.
Would appreciate some help with this question.
 A: $|f_n(x)|=|\frac{\sin(nx)+\cos(nx)}{\ln(n)}|
=
\sqrt{2} |\frac{ \sin(nx +\pi/4)}{\ln(n)}| \leq \sqrt{2} \frac{1}{\ln(n)}.$ The right side converges to $0,$ and this whole process is indepent of of the choice of $x.$
Edit: If you want to use the Cauchy criterion, the proof is essentially the same. Do this same business for both fractions. Without loss of generality, we can choose one guy that is bigger than another one: say, $ k>n.$ Then we've $$ |f_n(x) -f_k(x)|\leq \frac{2 \sqrt{2}}{\ln(n)}.$$ This shows that $f_n$ is Cauchy.
Extra: Note that $a \sin(x) +b \cos(x) \leq \sqrt{a^2 +b^2.}$
A: Since $|\sin u+\cos u|\le \sqrt{2}$, you can simply write
$$
\left|\frac{\sin( nx) +\cos( nx)}{\ln( n)} -\frac{\sin( kx) +\cos( kx)}{\ln( k)}\right|
{\le
\left|\frac{\sin( nx) +\cos( nx)}{\ln( n)}\right| +\left|\frac{\sin( kx) +\cos( kx)}{\ln( k)}\right|
\\\le
\frac{\sqrt 2}{\ln n}+\frac{\sqrt 2}{\ln k}
\\
<\epsilon,
}
$$
where the last inequality takes place for example when $n,k> \exp(\frac{3}{\epsilon})$.
