Here are some other problems concerning converging of sequences:
Suppose $p(x) = 2x^{4}+x^{2}+3x+1$ and $q(x) = 3x^{4}+x^{3}+2x+3$. Define $a_n = \frac{p(n)}{q(n)}$ for $n \in \mathbb{Z}^{+}$. Prove that $a_n \to \frac{2}{3}$.
In other words, I want to show that $(a_{n}-\frac{2}{3})$ is a null sequence. Now we know that $$a_n = \frac{ 2n^{4}+n^{2}+3n+1}{3n^{4}+n^{3}+2n+3}, \ n \in \mathbb{Z}^{+}$$
So $$a_n \leq \frac{2n^4+3n^2}{3n^4} = \frac{2}{3} + \frac{3}{n^2}$$
Thus $a_n \to \frac{2}{3}$. Is this the best way of doing the problem (i.e. it is bounded and monotonic increasing)?
Let $[a_n, b_n]$ be a nested sequence of closed intervals such that $b_n-a_n \downarrow 0$ and let $(x_n)$ be a sequence such that $x_n \in [a_n, b_n]$ for all $n$. Prove that $(x_n)$ is convergent.
We know that $$\bigcap (b_n-a_n) = 0$$ for all $n$. So I think $$\text{lim inf} \ \left( [a_n, b_n] \right) = \text{lim sup} \ \left([a_n, b_n] \right)$$
This is how I think of it: $(x_n)$ becomes "trapped" in smaller and smaller intervals until it is forced to converge to a point. Is this the right way to think about it?
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