Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\beta$.

share|cite|improve this question
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
up vote 2 down vote accepted

$n\beta-1<[n\beta]\leq n\beta$ 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 $\frac{1}{\beta}>1$.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.