How to prove that a sequence is not converget just using the definition? Let $X_n = \frac {3n^2}{2n-20}$. Prove if it is convergent or not using the definition.
I know $X_n$ is not convergent and I think that it is easier to prove that is unbounded, so the sequence is not convergent.
How can I prove it just using the definition?
Do I find an $\epsilon > 0$ such that $\forall N $  if $ n>N$ then $|X_n - L|>\epsilon$ ? I am confused by using the negative of the definition
 A: Let $N'=11$.  Then for $n\ge N'$ we have
$$\frac{3n^2}{2n-20}>\frac{3n^2}{2n}=\frac{3n}2$$
Therefore, for any given number $\epsilon>0$, we have 
$$\frac{3n^2}{2n-20}>\epsilon$$
whenever $n>\max\left(N',\frac23\epsilon\right)$.  And we are done!
A: If $x_n$ is convergent then
$$\exists\ell\quad\forall\epsilon>0\quad\exists N\quad\forall n\ge N\quad |x_n-\ell|<\epsilon.$$
The negation of this is:
$$\forall\ell\quad\exists\epsilon>0\quad\forall N\quad\exists n\ge N \quad|x_n-\ell|\ge\epsilon.$$
I hope you see the pattern of similarities between the two. To prove you can replace $\forall$ by "Let" and replace $\exists$ with a construction. The statement that $x_n$ is unbounded is:
$$\forall M\in\mathbb N\quad\forall N\quad\exists n\ge N \quad|x_n|\ge M.$$
Is this information sufficient to point you in the right direction?
A: One approach I see is the following:
Let $n >10$. We know $2n-20<2n\space$ so that implies $$\space\frac{1}{2n}<\frac{1}{2n-20} \implies \space\frac{3n^2}{2n} = \frac{3n}{2}<\frac{3n^2}{2n-20}$$ and since both quantities are positive, $$\left|\frac{3n}{2} \right| < \left|\frac{3n^2}{2n-20}\right| $$ for all $n>10$. Now show that $\lim_{n \to \infty} \frac{3n}{2}$ does not converge, and by comparison you can show $X_n$ does not converge.
