Determining if a sequence converges and proving it

Question

I have this sequence: $X_n := \frac {n^2 + 9n}{4n^4 + 4} $

I have to decide if it converges or diverges, and have to prove it. This is what I tried:

I think it converges to 0 because the power of $n$ in the denominator is a lot bigger than in the numeration ($4n^4$ against $n^2$ and $9n$).

Now I had to prove that, I based my proof on this video.

Proof: $x_{n}$ can only be divergent if for all epsilons bigger than zero, there exists some M bigger than zero, so that if n is bigger than M, $ | a_{n} - 0 | < \epsilon $.

$ | a_{n} - 0 | < \epsilon \\ |a_{n}| < \epsilon \\ |\frac{n^2 + 9n}{4n^4 + 4}| < \epsilon \\ \frac{n^2 + 9n}{4n^4 + 4} < \epsilon \\ \frac{4n^4 + 4}{n^2 + 9n} > \frac{1}{\epsilon} \\ \blacksquare $

Because $ M = \frac{1}{\epsilon} \\ n > \frac{1}{\epsilon} => |a_{n} - 0| < \epsilon$

I think this is pretty correct. Because my equations are true for any n bigger than zero I fill inm but I am not sure and I have trouble deciding if a sequence converges or diverges.

This is the rule I followed to decide that:

If the degree of the numerator is the same as the degree of the denominator, then the sequence converges to the ratio of the leading coefficients. If the denominator has a higher degree, then the sequence converges to 0; if the numerator has a higher degree, then the sequence diverges to ∞ if the leading coefficients have the same sign, or to −∞ if they have different signs.

Is this rule always correct? Is my proof correct? Any tips or suggestions on how to improve it or do it better?

Answer 1

or you write $$\frac{1}{n^2}\cdot \frac{1+\frac{9}{n}}{4+\frac{4}{n^4}}$$ and this tends to Zero if $n$ tends to infinity

Answer 2

The rule is correct, and to see it just divide everything by the highest degree of $n$. For instance, in your case

$$\frac{n^2+9n}{4n^4+4} = \frac{n^4}{n^4} \cdot \frac{\frac{1}{n^2}+\frac{9}{n^3}}{4 + \frac{4}{n^4}} = \frac{\frac{1}{n^2}+\frac{9}{n^3}}{4 + \frac{4}{n^4}} \underset{n\to \infty}{\longrightarrow} \frac{0 + 0}{4 + 0} = 0.$$

But watch out: your proof with the $\varepsilon$ criterion is not properly spelled out, you never say anything about your specific sequence. What would be different in the proof if your sequence were divergent? Moreover, you have written "$x_n$ can only be divergent..." instead of "$x_n$ can only be convergent to $0$..."

Answer 3

You probably want to write something like

$$\frac{n^2+9n}{4n^4+4} < \frac{n^2+9n^2}{4n^4} = \frac{5}{2n^2}.$$

You want this last bit to be less than $\epsilon,$ so you conclude that you want

$$n >\sqrt{\frac{5\epsilon}{3}}.$$

So choose $M$ bigger than $\sqrt{5\epsilon/3}$. Then if $n>M$ the above calculations show that $a_n < \epsilon.$