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I need to prove that $a_n = \frac{n^2+3n+2}{6n^3+5}$ converges as $n\to\infty$. I think it will converge to $0$, so I want to prove it.

Let $\epsilon >0$. Then

side-work $$\left|\frac{n^2+3n+2}{6n^3+5}\right|< \left| \frac{n^2+3n+2n}{6n^3+5}\right| < \left| \frac{n^2+5n}{6n^3}\right|$$ where I used $n \geq 1$ and then I continued $$\left| \frac{n^2+5n}{6n^3}\right| = \left|\frac{n+5}{6n^2}\right|<\left|\frac{6n}{6n^2}\right|$$ using again $n \geq 1$ and finally $$\left|\frac{6n}{6n^2}\right| = \left|\frac{1}{n}\right|=\frac{1}{n} < \epsilon$$

..Then take $n_0 = \max\left\{1, \frac{1}{\epsilon}\right\}$ and $\forall n > n_0$ we have $\left|\frac{n^2+3n+2}{6n^3+5}-0\right|<\epsilon$

Is this proof correct? Because my books give a different one

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    It's perfectly fine – 2017-01-23
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    $\left|\frac{6n}{6n^2}\right| = \left|\frac{1}{n}\right|=\frac{1}{n}$ – 2017-01-23
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    @manhattan thank you, long nights make this tricks – 2017-01-23
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    $$ \begin{align} \lim_{n\rightarrow\infty}a_n &=\lim_{n\rightarrow\infty}\frac{n^2+3n+2}{6n^3+5} =\frac{\infty}{\infty}\space\{\text{L'Hopital's rule}\}\space =\lim_{n\rightarrow\infty}\frac{\frac{d}{dn}\left(n^2+3n+2\right)}{\frac{d}{dn}\left(6n^3+5\right)} \\ &=\lim_{n\rightarrow\infty}\frac{2n+3}{\quad 18n^2 \quad} \quad =\frac{\infty}{\infty}\space\{\text{L'Hopital's rule}\}\space =\lim_{n\rightarrow\infty}\frac{\frac{d}{dn}\left(2n+3\right)}{\frac{d}{dn}\left(18n^2\right)} \\ &=\lim_{n\rightarrow\infty}\frac{2}{36n} =\color{red}{0} \end{align} $$ – 2017-01-23

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Unless you have some particular reason for an epsilon-delta proof:

$$\frac{n^2+3n+2}{6n^3+5} = \frac{n^2}{n^3} \cdot \frac{1 + 3/n + 2/n^2}{6+5/n^3} \;\;\to\;\; 0 \cdot \frac{1}{6} \;=\; 0$$

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    thank you! Do you reckon that one should prove a sequence converges, from the definition only when required to do so? What I mean is, is it considered acceptable at high rigour mathematics to provide such computation? – 2017-01-23
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    @Euler_Salter It depends on what theorems you've been shown in class. – 2017-01-23
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    @Euler_Salter It is perfectly acceptable (sometimes even required) to prove *anything* from first principles (only, you didn't state such requirement in the question). It is also perfectly acceptable to use previously proved results as "shortcuts" to simplify one's own proofs. For example, my answer is entirely rigorous provided that you already proved (or can otherwise assume) the common theorems about sums, products and ratios of convergent sequences. – 2017-01-23