Evaulate $ \lim_{n\to\infty} \frac{\sum_{k=1}^n\left[x^k\right]}{ x^n }$ I tried using Squeeze theorem :
$$x\le[x]\lt x+1$$
$$x^2\le[x^2]\lt x^2+1 $$
$$\vdots$$
$$x^n\le[x^n]\lt x^n+1$$
Therefore :
$$\frac{x+x^2+\cdots+x^n}{x^n}\le\frac{[x]+[x^2]+\cdots+[x^n]}{ x^{n} }\lt \frac{x+x^2+\cdots+x^n}{x^n}+\frac{n}{x^n}$$
$${ \underbrace{\frac{x \times (1-x^n)}{x^n(1-x)}}_{L_1} } \le\frac{[x]+[x^2]+\cdots+[x^n]}{ x^{n} }\lt \underbrace{\frac{x \times (1-x^n)}{x^n(1-x)} + \frac{n}{x^n}}_{L_2} $$
$L_1=\cfrac{x- x^{n+1}}{x^n-x^{n+1}} \times \cfrac{x^{-n}}{x^{-n}}= \cfrac{x^{1-n}-x}{1-x}$  and  $L_2=L_1 + \cfrac{n}{x^n}$
For ${ \left| x \right| } \gt 1$ , ${ \lim_{n\to\infty} L_1 }={ \lim_{n\to\infty} L_2 }=\frac{x}{x-1}$
Hence we conclude that: ${ \lim_{n\to\infty} \frac{[x]+[x^2]+\cdots+[x^n]}{ x^{n} } }= \frac{x}{x-1}$ for  ${ \left| x \right| } \gt 1$
And it is obvious  ${ \lim_{n\to\infty} \frac{[x]+[x^2]+\cdots+[x^n]}{ x^{n} } }= n$ for $x=1$ .
But I don't know how to evaluate the limit for $-1\lt x \lt 0$ and $0 \lt x \lt 1$ and $x=-1$

Hint:

*

*For $0<x<1$, $[x^n]=0$ for all $n\geq1$


*For $-1<x<0$, $[x^n]=-1$ if $n$ is odd and $0$ otherwise.


*For $x=1$ the expression is $n$


*For $x=-1$ the expression is $0$ if $n$ is even and $1$ of $n$ is odd.
