Is the limit of sequence enough of a proof for convergence? I have a sequence $a_{n}=\frac{(n+1)(n^{2})}{(2n+1)(3n^{2}+1)}$ ,
and limit of it when $n$ goes to infinity is $\frac{1}{6}$.
Because limit is some number, is that enough of a proof that this sequence is convergent, or should i do something more?
 A: Theorems on limits lift up the burden of doing $\varepsilon$-$\delta$ proofs. If you know that $\lim_{n\to\infty}a_n=l$ and $\lim_{n\to\infty}b_n=m$ (real $l$ and $m$), then you can also say
\begin{gather}
\lim_{n\to\infty}(a_n+b_n)=l+m \tag{Theorem 1}\\[6px]
\lim_{n\to\infty}(a_nb_n)=lm \tag{Theorem 2}\\[6px]
\lim_{n\to\infty}\frac{a_n}{b_n}=\frac{l}{m} \tag{Theorem 3}
\end{gather}
Theorem 3 requires the hypothesis $m\ne0$ (so also $b_n\ne0$ from some point on).
Such theorems can also be extended to the cases when one or both the limits are infinity, but this would take too far.
In your case, the sequence $a_n$ is not, as written, in one of the above forms, but you can rewrite it as
$$
a_n=\frac{n+1}{2n+1}\frac{n^2}{3n^2+1}
$$
Now let's examine
$$
b_n=\frac{n+1}{2n+1}
$$
We can't apply the third theorem above, but as soon as we rewrite
$$
b_n=\frac{1+\frac{1}{n}}{2+\frac{1}{n}}
$$
we see that at numerator and denominator we have sequences to which we can apply the first theorem above, because we know that $\lim_{n\to\infty}\frac{1}{n}=0$. Thus, combining theorems 1 and 3, we get
$$
\lim_{n\to\infty}b_n=\frac{1}{2}
$$
Similarly
$$
\lim_{n\to\infty}\frac{n^2}{3n^2+1}=
\lim_{n\to\infty}\frac{1}{3+\frac{1}{n^2}}=\frac{1}{3}
$$
Now apply theorem 2 and the requested limit is $\frac{1}{6}$.
More simply, you can directly do as in Olivier Oloa's answer. Mentioning the application of the above theorems is usually omitted (like Olivier did).
You don't need to check the limit with an $\varepsilon$-$\delta$ proof, because you're applying theorems that have been proved correct and the known fact that $\lim_{n\to\infty}\frac{1}{n}=0$.
A: It is enough to write, for $n>1$,
$$
a_{n}=\frac{(n+1)(n^{2})}{(2n+1)(3n^{2}+1)}=\frac16\cdot\frac{1+\frac1n}{\left(1+\frac1{2n}\right)\left(1+\frac1{3n^2}\right)}
$$ giving, as $n \to \infty$,
$$
a_n \to \frac16\cdot\frac{1+0}{\left(1+0\right)\left(1+0\right)}=\frac16.
$$
A: I will just elaborate on Olivier Oloa's answer. In the comments you mention you are unsure whether you should use "that $\epsilon$ proof". You don't have to (but sort of are anyway) and here's why.
The definition of $\lim_{n\to \infty} a_n = a$ is something like this
$$\forall \epsilon>0 \;\exists n_0\in \mathbb N,\; \forall n \geq n_0: |a_n - a| < \epsilon$$
Now, if you had just that and nothing else, you would indeed need to prove that your sequences converges to $1/6$ using the definition, i.e. "that $\epsilon$ proof"
Luckily, you've probably been shown (or even have proved) some basic results, namely things such as
$$\lim_{n\to \infty} \frac 1n = 0$$
and (for $\lim a_n = a$ and $\lim b_n = b$)
$$\lim_{n \to \infty} a_n + b_n = a+b$$
and so on.
So altogether what you're doing in the answer you've given is repeatedly using all these already proven theorems and rules to prove what your limit goes to (I recommend very carefully going through your argument to see what exact rules you've used). The theorems themselves are proven by the whole $\epsilon$ machinery, giving you the power to prove the convergence of the sequence without using a single $\epsilon$
Example:
Prove that
$$\lim_{n\to\infty} \frac 2n = 0$$
Now, you could do this (and quite easily) through the definition, i.e. doing the $\epsilon$ proof. But, using the two facts I wrote down earlier, you could argue
$$\lim_{n\to\infty} \frac 2n = \lim_{n\to\infty} \frac 1n + \frac 1n = \lim_{n\to\infty} \frac 1n + \lim_{n\to\infty} \frac 1n = 0 + 0 = 0$$
While this might be a quite convoluted usage (what rule could you use to prove this perhaps more directly?), it illustrates how one escapes the need to prove things from definition by using already proven facts/theorems.
