A few questions about trigonometric functions.

1. Set constant $a_{1},a_{2},\cdots,a_{n}$ meet $a_{1}+a_{2}+\cdots+a_{n}=0$, verify:

$$\lim_{x\to \infty}\sum_{k=1}^{n}a_{k}\sin \sqrt{x+k}$$

Proof. First, the conclusion of easy evidence

$$\lim_{x \to \infty}\sin \sqrt{x+k}-\sin \sqrt{x+n}=0$$

_{n}=-a_{1}-a_{2}-\cdots-a_{n-1}$ the $a into $\sum_{k=1}^{n}a_{k}\sin \sqrt{x+k}$.

$$\lim_{x\to \infty}\sum_{k=1}^{n}a_{k}\sin \sqrt{x+k}=\lim_{x\to \infty}\sum_{k=1}^{n-1}a_{k} (\sin \sqrt{x+k}-\ Sin \sqrt{x+n}) =0$$

The certificate is completed.

2. If you have $|\sum_{k=1}^{n}a_{k}\sin for any $x \in ( -1,1) $ kx| \leq |\sin x|$, verify:

$$|\sum_{k=1}^{n}k a_{k}| \leq 1$$

Proof: We use mathematical inductive method to prove.

(i). When $n=1$, it is easy to prove that the conclusion is established.

(ii). It is advisable to set up $n$ case proposition, to deduce the $n+1$ situation, if

$$|\sum_{k=1}^{n+1}a_{k}\sin kx|=|\sum_{k=1}^{n}a_{k}\sin kx +a_{n+1}\sin Nx\cos x+a_{n+1}\cos NX \sin X|\leq |\sin x|$$

Known by triangular inequalities

$$|\sum_{k=1}^{n-1}a_{k}\sin kx+ (A_{n}+a_{n+1}\cos x) \sin Nx|\leq (1+|a_{n+1}\cos nx|) | \sin x|$$

That

$$\sum_{k=1}^{n-1}\frac{a_{k}}{1+|a_{n+1}\cos nx|} \sin kx+\frac{a_{n}+a_{n+1}\cos X}{1+|a_{n+1}\cos x|} \sin Nx|\leq |\sin x|$$

It is understood by inductive method

$$\sum_{k=1}^{n-1}\frac{k A_{k}}{1+|a_{n+1}\cos nx|} +\frac{n (a_{n}+a_{n+1}) \cos X}{1+|a_{n+1}\cos x|}| \leq 1$$

Thus

$$|\sum_{k=1}^{n-1}k a_{k}+n a_{n}+n a_{n+1}\cos x|\leq 1+|a_{n+1}\cos nx|$$

By the triangular inequalities.

$$|\sum_{k=1}^{n+1}k A_{k}|\leq 1+|a_{n+1}\cos nx|-|n\cos x (n+1) | | a_{n+1}|$$

Make $x=0$, get

$$|\sum_{k=1}^{n+1}k A_{k}|\leq 1$$

3. $n$ of arbitrary positive integers, proving: when $x\in (0,\PI) $, constant

$$\sum_{k=1}^{n}\frac{\sin kx}{k}>0$$

A few questions about trigonometric functions.