Prove that $\inf\limits_{n\in\mathbb N}\sum\limits_{k=0}^{p}\lvert\sin{(n+k)^p}\rvert>0$

For any positive integer number $p$, show that $$\inf\left\{ {\left\vert\sin{(n^p)}\right\vert+\left\vert\sin{(n+1)^p}\right\vert+\cdots+ \left\vert\,\sin{(n+p)^p}\right\vert\, :\,n\in \mathbb{N}}\right\}>0. $$

My try. I only can prove for $p=2$. I have proved the following $$\vert\sin{(n+1)^2}\vert+\vert\sin{(n-1)^2}\vert+\vert\sin{n^2}\vert\ge\dfrac{1}{2}\sin{2}$$

Lemma. for any $x,y$ we have $$ \lvert\sin{x}\rvert+\lvert\sin{y}\rvert\ge\lvert\sin{(x-y)}\rvert. $$

This lemma is easy to prove.

Proof: \begin{align*} &|\sin{(n+1)^2}|+|\sin{(n-1)^2}|+|\sin{n^2}|\ge\dfrac{1}{2}[\sin{(n+1)^2}|+|\sin{(n-1)^2}|+2|\sin{n^2}|]\\ &\ge\dfrac{1}{2}(|\sin{[(n+1)^2-n^2]}|+|\sin{[n^2-(n-1)^2]}|)\\ &\ge\dfrac{1}{2}|\sin{[(n+1)^2+(n-1)^2-2n^2]}|=\dfrac{1}{2}\sin{2} \end{align*}

But my problem I can't it,

This problem is from a analysis problem book. This author's only hint, we note this $$\sum_{i=0}^{p}(-1)^i\binom{p}{i}(n+p-i)^p=p!$$ But I can't. Thank you

Solution 1:

Let's assume that $$ \inf_{n\in\mathbb N}\lvert\sin{(n^p)}\rvert +\lvert\sin{(n+1)^p}\rvert+\cdots+\lvert\sin{(n+p)^p}\rvert=0. $$ Then for every $\varepsilon>0$, then exists an $n$, such that: $\lvert\sin{(n+j)^p}\rvert<\varepsilon/2$, for every $j=0,1,\ldots,p$.

But $\lvert\sin x\rvert<\varepsilon$ implies that there exists an $m\in\mathbb Z$, such that $\lvert x-m\pi\rvert<\varepsilon$.

Hence there exist $m_j\in\mathbb N$, such that $$ \lvert m_j\pi-(n+j)^p\rvert<\varepsilon, \quad\text{for every}\,\,\, j=0,1,\ldots,p.\tag{1} $$ or equivalently $$ (n+j)^p=m_j\pi+\delta_j, \quad\text{with}\,\,\, \lvert\delta_j\rvert<\varepsilon. \tag{1'} $$ We shall use the following identity (which was a hint given to the OP): $$ \sum_{k=0}^p(-1)^k\binom{p}{k}(n+p-k)^p=p!\tag{2} $$ See formula (10) in Binomial Coefficients. Combination of $(1')$ and $(2)$ provides $$ p!=\sum_{k=0}^p(-1)^k\binom{p}{k}(m_j\pi+\delta_j)=N\pi+\pi\sum_{k=0}^p(-1)^k\binom{p}{k}\delta_j $$ where $N=\sum_{k=0}^p(-1)^k\binom{p}{k}m_j$ is an integer. Thus $$ |p!-N\pi|\le \pi\sum_{k=0}^p\binom{p}{k}|\delta_j|\le \varepsilon\pi\sum_{k=0}^p\binom{p}{k} =\pi 2^p\varepsilon. \tag{3} $$ This is a contradiction as the left hand side can not be too small for the different values of $n$, as $\pi$ is irrational. In fact $$ |p!-N\pi|\ge \min\left\{p!-\pi\left\lfloor\frac{p!}{\pi}\right\rfloor, \pi\left\lfloor\frac{p!}{\pi}\right\rfloor+\pi-p!\right\}>0. $$ On the other hand, having assumed $(1)$, the left hand side of $(3)$ can become arbitrarily small.

This is a contradiction, and hence $$ \inf_{n\in\mathbb N}\lvert\sin{(n^p)}\rvert+\lvert\sin{(n+1)^p}\rvert +\cdots+\lvert\sin{(n+p)^p}\rvert>0. $$ Ὅπερ ἔδει δεῖξαι.

Solution 2:

Well, the author almost solved the problem for you by giving the hint. Let me just add some details.

You would like to show that $\inf\{ |\sin n^p|+|\sin (n+1)^p|+...|\sin (n+p)^p|\}$ is bounded away from $0,$ in other words all terms $|\sin (n+i)^p|$ for $0\le i\le p$ cannot be too small simultaneously. This in turn, is equivalent to saying that all $(n+i)^p,$ $0\le i\le p$ cannot be to close to $\pi n_i$ simultaneously. But, as you indicated above, $$\sum_{i=0}^p(n+i-p)^p {n\choose i}=p!\in\mathbb{Z},$$

and $\inf_{n\in\mathbb{N}}|\pi n-p!|>0$ which implies the result.

Solution 3:

An another proof if someone is interested,

• $(X^p,..., (X+p)^p)$ is a base of ${\mathbb{Q}}_p [X]$: There exist $m\in \mathbb{N^*}$ and $\lambda_0 , ... , \lambda_p \in \mathbb{Z^{p+1}}$ such that $m = \lambda_0 X^p + \lambda_1 (X+1)^p + ... + \lambda_p (X+p)^p$.
• Let's assume that $$ \inf_{n\in\mathbb N}|\sin{(n^p)}|+|\sin{(n+1)^p}|+\cdots+|\sin{(n+p)^p}|=0. $$ Then there exist a sequence $(n_k)$ such that $\forall j\in [[0,p]]$, $$\lim_{k\to +\infty}\sin ((n_k+j)^p )=0$$ i.e.
  $$\lim_{k\to +\infty} e^{2i(n_k+j)^p}=1$$ Thus, $e^{2im}$ tends to 1

Therefore $m=0$. This is a contradiction, and hence $$ \inf_{n\in\mathbb N}|\sin{(n^p)}|+|\sin{(n+1)^p}|+\cdots+|\sin{(n+p)^p}|>0. $$ QED