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How to show that $\lim_{n \rightarrow \infty} \frac{[a^{n+1}]}{[a^n]}=a$, where $[a]$ = integer part of a?
Here $a>1$. But I suspect it is true for all $a \ne 0$.

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    Oh, yes, I will correct it. Thanks. – 2017-01-04
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    The limit is undefined for all $0\le a<1$. – 2017-01-04
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    Simple and direct: $$a^{n+1}\leqslant [a^{n+1}]$$\frac{a^{n+1}}{a^n+1}\leqslant\frac{[a^{n+1}]}{[a^n]}\leqslant\frac{a^{n+1}+1}{a^n}$$ Can you finish? – 2017-01-04
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    For large numbers, $x$ and $[x]$ are virtually equal, hence $ a^{n+1}/a^n\to a$. – 2017-01-04
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    @Did I get the idea.(The relation should be the other way around). – 2017-01-04

2 Answers 2

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For $|a|>1$,

$$\frac{[a^{n+1}]}{[a^n]}=\frac{a^{n+1}-\{a^{n+1}\}}{a^n-\{a^n\}}=a\frac{1-\dfrac{\{a^{n+1}\}}{a^{n+1}}}{1-\dfrac{\{a^{n}\}}{a^n}}\to a.$$

As the fractional parts are bounded, the numerator and denominator both tend to $1$.

This can be extended to $|a|\ge1$ as with $|a|=1$, the fractional parts are all $0$.

For $0\le a<1$, the limit is undefined (none of the ratios are defined).

For $-1

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For $x>1$, we have the trivial inequalities $0

$\frac{a^{n+1}-1}{a^n}\leq \frac{[a^{n+1}]}{[a^n]}\leq \frac{a^{n+1}}{a^n-1}$

But it is easy to check that for $a>1$, we have:

$\frac{a^{n+1}-1}{a^n}=a-a^{-n}\rightarrow a-0=a$,

$\frac{a^{n+1}}{a^n-1}=\frac{a}{1-a^{-n}}\rightarrow \frac{a}{1-0}=a$

So finally by the squeeze theorem we conclude that $\frac{[a^{n+1}]}{[a^n]}\rightarrow a$

Note: Work similarly for the other values of $a$, but be carefull with the signs and don't expect always the same value.