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?

  • $\begingroup$ From the fact that $a_n \leq \frac{2}{3} + \frac{3}{n^2}$ you cannot conclude that $a_n\to \frac{2}{3}$; you can only conclude that, if the $a_n$ have a limit, then this limit is less than or equal to $\lim(\frac{2}{3}+\frac{3}{n^2}) = \frac{2}{3}$. That is, your conclusion does not follow from what you've written. $\endgroup$ – Arturo Magidin Jun 23 '11 at 21:06
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    $\begingroup$ Please stop putting multiple, unrelated questions into the same post. $\endgroup$ – Arturo Magidin Jun 23 '11 at 21:06
  • $\begingroup$ You have $a_n\leq x_n \leq b_n$; the $a_n$ are nondecreasing and bounded, the $b_n$ are nonincreasing and bounded, so they all converge. Therefore, if the $x_n$ converge, then $\lim a_n \leq \lim x_n \leq \lim b_n$. You know $\lim a_n = \lim b_n$ (since $\lim(b_n-a_n) = 0$); use that to show $(x_n)$ is Cauchy and hence convergent. P.S. $\cap (b_n-a_n)$? What does it mean to intersect a sequence of numbers? $\endgroup$ – Arturo Magidin Jun 23 '11 at 21:09
  • $\begingroup$ Yes, please stop posting multiple unrelated questions in same post. For instance, if it weren't for the second problem, this would be a dupe of: math.stackexchange.com/questions/33970/… $\endgroup$ – Aryabhata Jun 23 '11 at 21:16
  • $\begingroup$ Yes, as Arturo points out about your 2nd question (intersecting sequences?): it would be correct to note that the intersection of all such nested intervals is the set containing a single point: i.e. ${0}$. $\endgroup$ – Namaste Jun 23 '11 at 21:51

For limits of the kind $\lim_{x \to \infty}\frac{f(x)}{g(x)}$ ($f,g$ polynomials) there is an algorithm. You need to know that $\lim_{x \to \infty}\frac{1}{x^k}=0,\ k >0$, so force out the greatest power of $x$ you can get from $f,g$.

Example in your case:


After simplifying the expressions you get some simple limits. If the degrees of polynomials are not the same, the result is $0$ if the greatest degree is in the denominator, or $\infty$ if it is the other way around.


Concerning the second question, as Arturo pointed out in a comment above, $\lim _{n \to \infty } a_n = \lim _{n \to \infty } b_n $. Denote the common limit by $l$. Since $a_n \leq x_n \leq b_n$ for all $n$, by the squeeze theorem $\lim _{n \to \infty } x_n = l$.

Also, with regard to amWhy's comment above, it is worth recalling the Nested Interval Theorem (though you don't need it for the present question).


Regarding the limit: Please see this FAQ thread: Finding the limit of $\frac{Q(n)}{P(n)}$ where $Q$ and $P$ are polynomials.

It shows how to find $$\lim_{n\rightarrow\infty}\frac{Q(n)}{P(n)}$$ for any polynomials $P,Q$. (Which is summarized in Beni Bogosel's answer)


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