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Bounded sequence

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Bounded sequence
Let \(\displaystyle{x\in\mathbb{R}}\). Prove that the sequence
\(\displaystyle{s_n=\sin\,x+...+\sin\,(n\,x)\,,n\in\mathbb{N}}\) is bounded.
\(\displaystyle{s_n=\sin\,x+...+\sin\,(n\,x)\,,n\in\mathbb{N}}\) is bounded.
 Grigorios Kostakos
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Re: Bounded sequence
For $x\neq2m\pi\,,\; m\in\mathbb{Z}$, maybe the formula
\[s_n=\mathop{\sum}\limits_{k=1}^n\sin(kx)=\frac{\sin\big(\frac{(n+1)\,x}{2}\big)\,\sin\big(\frac{nx}{2}\big)}{\sin\big(\frac{x}{2}\big)}\] and that \[s_n=\Bigg\frac{\sin\big(\frac{(n+1)\,x}{2}\big)\,\sin\big(\frac{nx}{2}\big)}{\sin\big(\frac{x}{2}\big)}\Bigg\leqslant\frac{1}{\big\sin\big(\frac{x}{2}\big)\big}\] could be useful.
edit: 28/10/2016, 07:40. The condition $x\neq0$ replased by $x\neq2m\pi\,,\; m\in\mathbb{Z}$.
\[s_n=\mathop{\sum}\limits_{k=1}^n\sin(kx)=\frac{\sin\big(\frac{(n+1)\,x}{2}\big)\,\sin\big(\frac{nx}{2}\big)}{\sin\big(\frac{x}{2}\big)}\] and that \[s_n=\Bigg\frac{\sin\big(\frac{(n+1)\,x}{2}\big)\,\sin\big(\frac{nx}{2}\big)}{\sin\big(\frac{x}{2}\big)}\Bigg\leqslant\frac{1}{\big\sin\big(\frac{x}{2}\big)\big}\] could be useful.
edit: 28/10/2016, 07:40. The condition $x\neq0$ replased by $x\neq2m\pi\,,\; m\in\mathbb{Z}$.
Last edited by Grigorios Kostakos on Fri Oct 28, 2016 5:41 am, edited 1 time in total.
Grigorios Kostakos
Re: Bounded sequence
What Grigorios has written is the one way to go I think!
Let us see something interesting here. Let us see what happens to infinity. We have seen that $s_n$ is bounded. Note that there are some values of $x$ for which the series $\sum \limits_{n=1}^{\infty} \sin nx$ converges and others that does not.
Let us see something interesting here. Let us see what happens to infinity. We have seen that $s_n$ is bounded. Note that there are some values of $x$ for which the series $\sum \limits_{n=1}^{\infty} \sin nx$ converges and others that does not.
 In a general sense , including Cesaro summation sense and Abel summation sense the following holds:
$$\sum_{n=1}^{\infty} \sin nx = \Im \left(\frac{e^{ix}}{1e^{ix}}\right) = \frac{1}{2}\cot\frac{x}{2}$$  The series converges if we assume that we are in the upper half plane , that is if $\Im (x) >0$, diverges if $x$ is real unless $x \in \pi \mathbb{Z}$.
 In general
$$\sum_{n = 1}^\infty \sin nx \neq \Im \bigg[\sum_{n = 1}^\infty e^{inx}\bigg]$$
Why?
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1p^{s}}$

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Re: Bounded sequence
It is not bounded near zero.
Re: Bounded sequence
Yes, you are right. I , myself, oversaw that. Thus we need to compe up with another stategy to attack this.S.F.Papadopoulos wrote:It is not bounded near zero.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1p^{s}}$

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Re: Bounded sequence
Here is a solution.
If \(\displaystyle{x=0}\) or \(\displaystyle{x=2\,\pi}\), then \(\displaystyle{s_n=0\,,n\in\mathbb{N}}\)
and we are done.
It is sufficient to prove the argument if \(\displaystyle{x\in\left(0,2\,\pi\right)}\) since the \(\displaystyle{\sin}\)  function
is a \(\displaystyle{2\,\pi}\)  periodical function.
So, for every \(\displaystyle{n\in\mathbb{N}}\), we have that
\(\displaystyle{\begin{aligned} s_n&=\sum_{k=1}^{n}\sin\,(k\,x)\\&=\dfrac{1}{2\,i}\,\sum_{k=1}^{n}\left(e^{i\,k\,x}e^{i\,k\,x}\right)\\&=\dfrac{1}{2\,i}\,\sum_{k=1}^{n}(e^{i\,x})^k\dfrac{1}{2\,i}\,\sum_{k=1}^{n}(e^{i\,x})^k\,\,(I)\end{aligned}}\).
Since \(\displaystyle{x\in\left(0,2\,\pi\right)}\), we get \(\displaystyle{e^{i\,x}\,,e^{i\,x}\neq 1}\),
so the relation \(\displaystyle{(I)}\) gives us
\(\displaystyle{\begin{aligned} s_n&=\dfrac{e^{i\,x}}{2\,i}\,\dfrac{e^{i\,n\,x}1}{e^{i\,x}1}\dfrac{e^{i\,x}}{2\,i}\,\dfrac{e^{i\,n\,x}1}{e^{i\,x}1}\\&=\dfrac{e^{i\,x}}{2\,i}\,\dfrac{e^{i\,n\,x}1}{e^{i\,x}1}+\dfrac{1}{2\,i}\,\dfrac{e^{i\,n\,x}1}{e^{i\,x}1}\end{aligned}}\)
Therefore,
\(\displaystyle{s_n\leq \dfrac{2}{2\,e^{i\,x}1}+\dfrac{2}{2\,e^{i\,x}1}=\dfrac{1}{\left\sin\,\dfrac{x}{2}\right}\,,\forall\,n\in\mathbb{N}}\).
If \(\displaystyle{x=0}\) or \(\displaystyle{x=2\,\pi}\), then \(\displaystyle{s_n=0\,,n\in\mathbb{N}}\)
and we are done.
It is sufficient to prove the argument if \(\displaystyle{x\in\left(0,2\,\pi\right)}\) since the \(\displaystyle{\sin}\)  function
is a \(\displaystyle{2\,\pi}\)  periodical function.
So, for every \(\displaystyle{n\in\mathbb{N}}\), we have that
\(\displaystyle{\begin{aligned} s_n&=\sum_{k=1}^{n}\sin\,(k\,x)\\&=\dfrac{1}{2\,i}\,\sum_{k=1}^{n}\left(e^{i\,k\,x}e^{i\,k\,x}\right)\\&=\dfrac{1}{2\,i}\,\sum_{k=1}^{n}(e^{i\,x})^k\dfrac{1}{2\,i}\,\sum_{k=1}^{n}(e^{i\,x})^k\,\,(I)\end{aligned}}\).
Since \(\displaystyle{x\in\left(0,2\,\pi\right)}\), we get \(\displaystyle{e^{i\,x}\,,e^{i\,x}\neq 1}\),
so the relation \(\displaystyle{(I)}\) gives us
\(\displaystyle{\begin{aligned} s_n&=\dfrac{e^{i\,x}}{2\,i}\,\dfrac{e^{i\,n\,x}1}{e^{i\,x}1}\dfrac{e^{i\,x}}{2\,i}\,\dfrac{e^{i\,n\,x}1}{e^{i\,x}1}\\&=\dfrac{e^{i\,x}}{2\,i}\,\dfrac{e^{i\,n\,x}1}{e^{i\,x}1}+\dfrac{1}{2\,i}\,\dfrac{e^{i\,n\,x}1}{e^{i\,x}1}\end{aligned}}\)
Therefore,
\(\displaystyle{s_n\leq \dfrac{2}{2\,e^{i\,x}1}+\dfrac{2}{2\,e^{i\,x}1}=\dfrac{1}{\left\sin\,\dfrac{x}{2}\right}\,,\forall\,n\in\mathbb{N}}\).
 Grigorios Kostakos
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Re: Bounded sequence
I think there is a misunderstanding here about boundedness. Let me try to elucidate. If we consider $x$ as a fixed real number, then the sequence $s_n(x)$ is a sequence of real numbers and a boundary for this sequence is $\frac{1}{\sin({x}/{2})}$.Papapetros Vaggelis wrote:Let \(\color{red}{x\in\mathbb{R}}\). Prove that the sequence
\(\displaystyle{s_n=\sin x+\dots+\sin(n\,x)\,,\; n\in\mathbb{N}}\) is bounded.
But, if we think $s_n(x)$ as an sequence of functions, then boundedness, means uniform boundedness, that is there exists a real number $M$ such that \[(\forall\, x\in\mathbb{R})(\forall\, n\in\mathbb{N}) \quad s_n(x)\leqslant M.\]
So, $s_n(x)$ is bounded for every $x\in\mathbb{R}$ by $\frac{1}{\sin({x}/{2})}$, but it is not uniformly bounded(*), according to the definition above.
(*) Proof left as an exercise!
P.S. I don't blame anyone here! I think that Vaggelis was quite clear by putting $x\in\mathbb{R}$ in the first line.
Grigorios Kostakos