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Problem: Show that the following series $\sum\limits_{n=1}^{\infty}\frac{\sin(nx)}{{(n^4+x^4)}^{\frac{1}{3}}}$ is convergent.

It was expressly indicated that I should prove the series convergence in the following way:

Dirichlet´s test for Uniform convergence: The series converge uniformly

$\sum_\limits{n=k}^{\infty}f_ng_n$

on $S$ if $\{f_n\}$ converges uniformly to zero on $S$, $\sum(f_{n+1}-f_n)$ converges absolutely uniformly on $S$, and

$||g_k+g_{k+1}+...||_S\leqslant M\:\:n\geqslant K$

for some constant $M$.

Resolution attempt:

$f_n=\frac{1}{{(n^4+x^4)}^{\frac{1}{3}}}\to 0$ as $n\to\infty$

Now I must prove $\sum\limits_{n=1}^{\infty}{\sin(nx)}\leqslant M$

$\sin(nx)=\frac{e^{inx}-e^{-inx}}{2i}$

Question:

How can I use $\frac{e^{inx}-e^{-inx}}{2i}$ to major $\sum\limits_{n=1}^{\infty}{\sin(nx)}$?

Thanks in advance!

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  • $\begingroup$ For which $x?\,$ $\endgroup$ – zhw. Mar 3 '18 at 17:04
  • $\begingroup$ @zhw. The problem intends to find the dominion of convergence. $\endgroup$ – Pedro Gomes Mar 3 '18 at 17:07
  • $\begingroup$ The problem as you stated it doesn't say that. It says to show the series is convergent. Is $x$ real, complex? Please make these things clear. $\endgroup$ – zhw. Mar 3 '18 at 17:27
  • $\begingroup$ @zhw. That is not defined by the author, so I assume that $x\in\mathbb{C}$. Can you work it out now? $\endgroup$ – Pedro Gomes Mar 3 '18 at 18:08
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    $\begingroup$ For real $x$ we have $ |(\sin nx)/(n^4+x^4)^{1/3}|\leq $ $1/(n^4+x^4)^{1/3}\leq $ $1/n^{4/3}$ and the series converges absolutely. For $x \in \Bbb C,$ if $x^4+n^4=0$ for some $n\in \Bbb N$ then one of the terms in the series does not exist. If $x^2+1=0$ then $|\sin nx| /(e^n/2)\to 1$ as $n\to \infty$ so the absolute values of the terms of the series tends to $\infty.$ And there are other $x\in \Bbb C$ for which the series diverges..... Also the set $\{\sum_{n=1}^k\sin nx\; :k\in \Bbb N\}$ is unbounded for some real $x.$ $\endgroup$ – DanielWainfleet Mar 4 '18 at 5:13
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From $\cos (A\pm B)=\cos A \cos B \mp \sin A \sin B$ we have $\cos (A-B)-\cos (A+B)=2\sin A \sin B.$

Therefore (with $h(x,n)$ being an abbreviation) we have $$[1]\quad 2\sin mx \sin (x/2)= h(x,m) =\cos (m-1/2)x-\cos (m+1/2)x. $$

Let $f(x,n)=\sum_{m=1}^n \sin mx.$ Then $$f(x,n)\sin (x/2)=\frac {1}{2}\sum_{m=1}^n h(x,m).$$ From the RHS of $[1]$ this is a telescoping sum, so $$f(x,n)\sin (x/2)=\cos (x/2)-\cos (n+1/2)x.$$ If $\sin (x/2)\ne 0 \;$ (that is, if $\frac {x}{2\pi} \not \in \Bbb Z $ ) then $$f(x,n)=\cot (x/2)-\frac {\cos (n+1/2)x} {\sin (x/2)}.$$ From this I see that my comment that $\{f(x,n): n\in \Bbb N\}$ may be unbounded for some real $x$ is false. (Sorry about that.).

However if $0\ne ix\in \Bbb R$ then $\cos (n+1/2)x=\cosh (n+1/2)x \to \infty$ as $n\to \infty.$

For the (trivial) case $\sin (x/2)=0$ we have $\sin nx=0$ for all $n\in \Bbb N$ and $f(x,n)=0.$

For $x\in \Bbb R$ we may obtain the closed formula for $f(x,n)$ by observing that it is the imaginary part of $g(x,n)=\sum_{m=1}^n e^{mix}.....$ If $e^{ix/2}\ne 1$ then $$g(x,n) =e^{ix}\cdot \frac {e^{nix}-1}{e^{ix}-1}=$$ $$=\frac {e^{ix/2}}{2i\sin (x/2)}\cdot e^{nix/2}\cdot 2i\sin (nx/2)=$$ $$=e^{i(n+1/2)x}\cdot\frac {\sin (nx/2)}{\sin (x/2)}.$$

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