Find $\mathop {\lim }\limits_{x \to \infty }(\sin x + \sin x^2+...+\sin x^n)$ Find $$\mathop {\lim }\limits_{x \to \infty }(\sin x + \sin x^2+...+\sin x^n)$$
I can't show that it doesn't exist. Thank for help me. 
 A: Regarding Dr. MV's comment, the terms are ambiguous as written. If your terms are $(\sin x)^i$ then you may consider using a geometric series identity to write the function as $\frac{1-(\sin x)^n}{1-\sin x}\sin x$, and noting that the function $\frac{(1-u^n)u}{1-u}$ is nonconstant on $[-1,1]$.
For me the more interesting case is when the terms are equal to $\sin(x^i)$ so I will cover that case.
Let $f_1(x)=\sin(x)+\sin(x^2)+...\sin(x^{n-1})$, and let $f_2(x)=\sin(x^n)$. Let $f=f_1+f_2$, so we want to show that $\lim_{x \to \infty} f(x)$ does not exist.
We define $x_k:=\big((k+1/2)\pi\big)^{1/n}$ for each $k \geq 0$.
First notice that $f_2(x_k)=(-1)^k$.
Next notice that for every $\epsilon>0$ there exists some $N$ such that if $k \geq N$ then $|f_1(x_{k+1})-f_1(x_k)|<\epsilon$. This follows from the fact that $\sin$ is uniformly continuous on $\Bbb R$ together with the fact that for $i<n$ the successive differences of the powers $x_{k+1}^i-x_k^i$ approach zero as $k \to \infty$.
This shows that $|f(x_{k+1})-f(x_k)| \geq |f_2(x_{k+1})-f_2(x_k)| - |f_1(x_{k+1})-f_1(x_k)| \geq 2-\epsilon$ as soon as $k \geq N$. Therefore $x_k \to \infty$ and $f(x_k)$ diverges, which shows that $f$ does not have an infinite limit.
A: Let $$f_n(x) = \sum_{k=1}^n \sin(x^k)$$
Note that we're keeping $n$ fixed. 
I claim that this sum diverges for all $n$ as $x \to +\infty$. I provide a semi-rigorous argument, which is not too difficult to make entirely rigorous. 
The idea is (roughly) the following. Let $\epsilon>0$ be small. If we make $x$ sufficiently large, there are intervals $[x, x+\delta]$ such that for $1\leq k<n$ we have $(x+\delta)^k - x^k < \frac{\epsilon}{n}$ but we have $(x+\delta)^n - x^n > 2\pi$. This is not difficult to prove. 
Therefore, by continuity, the lower order terms $\sin(x^k)$ for $1 \leq k < n$ do not vary too much in $[x, x+\delta]$. However, the higher order term $\sin (x^n)$ completes a full period, so to speak, in $[x, x+\delta]$ (e.g., it takes values from $-1$ to $1$). This implies that the function $f_n$ (modulo some small error) also completes a full "period" from $-1$ to $1$. As intervals of this sort are present arbitrarily far into the positive real line, this implies that $f_n(x)$ does not have a limit as $x \to \infty$.  
