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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.

Esty.R
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2 Answers2

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$|f_n(x)|=|\frac{\sin(nx)+\cos(nx)}{\ln(n)}| = \sqrt{2} |\frac{ \sin(nx +\pi/4)}{\ln(n)}| \leq \sqrt{2} \frac{1}{\ln(n)}.$ 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 $a \sin(x) +b \cos(x) \leq \sqrt{a^2 +b^2.}$

Matha Mota
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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 $n,k> \exp(\frac{3}{\epsilon})$.

Mostafa Ayaz
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  • Thank you for your answer. If you can please elaborate on the last inequality I don't see how if $\displaystyle n,k >e^{\frac{3}{\varepsilon }}$ solves the question. By plugging that into the inequality I get: $\displaystyle \frac{\sqrt{2}}{\frac{3}{\varepsilon }} +\frac{\sqrt{2}}{\frac{3}{\varepsilon }} =\frac{\sqrt{2} \cdotp \varepsilon }{3} +\frac{\sqrt{2} \cdotp \varepsilon }{3}$. Additionally I'm not sure this inequality holds for $\displaystyle n,k >e^{\frac{3}{\varepsilon }}$. I think it to should be the exact opposite - $\displaystyle n,k <e^{\frac{3}{\varepsilon }}$. – Esty.R May 19 '21 at 14:05
  • You're welcome. Sure I will elaborate. Generally in using $\epsilon-\delta$ definition of limits, you need not use strict bounds. Rather, sufficient conditions can help much. For example in this case, I have assumed that the conditions $\frac{\sqrt 2}{\ln n}<\frac{\epsilon}{2}$ and $\frac{\sqrt 2}{\ln k}<\frac{\epsilon}{2}$ are sufficient for deduction of $\frac{\sqrt 2}{\ln n}+\frac{\sqrt 2}{\ln k}<\epsilon$, both of which yielding $n,k>\exp(\frac{2\sqrt 2}{\epsilon})$. The final answer uses only a better notation since $3>2\sqrt{2}$. – Mostafa Ayaz May 19 '21 at 15:33
  • You're welcome. Sure I will elaborate. Generally in using $\epsilon-\delta$ definition of limits, you need not use strict bounds. Rather, sufficient conditions can help much. For example in this case, I have assumed that the conditions $\frac{\sqrt 2}{\ln n}<\frac{\epsilon}{2}$ and $\frac{\sqrt 2}{\ln k}<\frac{\epsilon}{2}$ are sufficient for deduction of $\frac{\sqrt 2}{\ln n}+\frac{\sqrt 2}{\ln k}<\epsilon$, both of which yielding $n,k>\exp(\frac{2\sqrt 2}{\epsilon})$. The final answer uses only a better notation since $3>2\sqrt{2}$. – Mostafa Ayaz May 19 '21 at 15:33
  • There is actually nothing special about $3$ as long as it is more than $2\sqrt 2$. – Mostafa Ayaz May 19 '21 at 15:33