Show that the sequence $\{a_n \}$ converges to $\beta$

For each $x \in \mathbb{R}$ let $[x]$ denote the greatest integer less than or equal to $x$. Further, for a fixed $\beta \in (0,1)$ define $a_n = (1/n) [n\beta] + n^2\beta^n$ for all $n \in \mathbb{N}$.Show that the sequence $\{ a_n \}$ converges to $\beta$.

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Hint: $n\beta-1\leq [n\beta]\leq n\beta$. Complete the inequality to have $a_{n}$ in the middle and conclude. – Lucien Sep 16 '12 at 8:08

$n\beta-1<[n\beta]\leq n\beta$ implies $\beta-\frac{1}{n}+\beta^nn^2\leq a_{n}\leq \beta +\beta^n n^2$. Take $n\to \infty$, and use that $\lim_{n\to\infty} \frac{n^2}{(1/\beta)^n}=0$ because $\frac{1}{\beta}>1$.

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The ratio $$\frac{ (n+1)^2 \beta^{n+1} }{n^2 \beta^n } = (1+1/n)^2 \beta\to \beta <1$$ so $n^2 \beta^n \to 0$ by comparison to a geometric sequence. Also, we have $x-1 \leq [x] \leq x$ so $$\frac{ n\beta -1}{n} \leq \frac{ [n\beta] }{n} \leq \frac{ n\beta }{n}$$

and so $\displaystyle \frac{ [n\beta] }{n}\to \beta$ by the Squeeze theorem.

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