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I am given the following succession/series/sequence:

$$ a_n = \frac{4n^5 +4n^3+n}{5n^4-2n^5+n^2} $$

How do I find out if it converges or diverges and how to find such values.

I am quite lost on the subject.

I've read that in a Geometric succession/series/sequence it is convergent if the ratio is less than 0, but I'm not sure if its a geometric series.

Help is really appreciated, thanks in advance.

PD: My native language is not english so I'm not sure what the appropriate term would be, is it succession, series or sequence.

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Terminology: if you are asking about $\lim_{n\to\infty}a_n$, that's a sequence. If you are asking about $\sum_{n=1}^{\infty}a_n$, that's a series. A geometric series converges if the common ratio is less than $1$ (not $0$) in absolute value. The sequence in this question is not geometric, and I'd recommend that you put some time into learning how to recognize a geometric progression when you see one, because that is a very useful skill to have. – Gerry Myerson Nov 25 '11 at 10:09

I assume you want the limit as $n\rightarrow\infty$.

You have a rational expression. The trick (or one trick) to find the limit is to divide every term by the term with the highest power: $n^5$.

This gives $$ {4n^5+4n^3+x\over -2n^5+5n^4+n^2} = {{4n^5\over n^5}+{4n^3\over n^5}+{n\over n^5}\over {-2n^5\over n^5}+{5n^4\over n^5}+{n^2\over n^5}} = {{4 }+{4 \over n^2}+{1\over n^4}\over {-2 }+{5 \over n }+{1\over n^3}}\ \buildrel{n\rightarrow\infty}\over{\longrightarrow}\ {4\over-2}=-2. $$

The above method can be used to establish rules given by Listing in his answer.

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There are three basic rules for evaluating limits at infinity for a rational function $f(x) = p(x)/q(x)$: (where $p$ and $q$ are polynomials):

  • If the degree of p is greater than the degree of q, then the limit is positive or negative infinity depending on the signs of the leading coefficients;

  • If the degree of p and q are equal, the limit is the leading coefficient of p divided by the leading coefficient of q;

  • If the degree of p is less than the degree of q, the limit is 0.

(Quoted from Wikipedia)

In your case you have

$p(x)=4x^5+4x^3+x$, $q(x)=-2x^5+5x^4+x^2$

It follows by point 2 that the lemit is equal to

$\lim_{x\rightarrow \infty}f(x)=\frac{4}{-2}=-2$

To formally prove those rules apply l'Hôpital $k$ times where $k$ is the lowest occuring degree in the fraction.

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I would say, to formally prove these rules, don't use l'Hopital. Instead, use the "trick" David Mitra uses in his answer. – Gerry Myerson Nov 25 '11 at 10:05
Yes, thats a more elementary way :) – Listing Nov 25 '11 at 10:07

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