If a series converges, then the sequence of terms converges to 0, Proof by contradiction. I know there is an elegant proof of this theorem,
But I've been ask to prove it in the test and the only thing I was thinking of was to prove it by contradiction.I want to know if I did right, or maybe I was assuming something wrong: 
${S_n} = \sum _{n=0}^{\infty }\:a_n$ converge and assume that $\lim _{x\to \infty }\left(a_n\right)\:=\:L\:\in \mathbb{R}$    (L isn't zero)
So there is $N\in \mathbb{N}\:\:\:\:\:\forall \:n>\:N\:\:\rightarrow \:a_n>\:L-\epsilon$
So if  $a_n>0 \Rightarrow \sum _{n=0}^{\infty }\:a_n>\sum _{n=0}^{\infty }\:L-\epsilon =\:\lim _{n\to \infty }\left(n\cdot \left(L-\epsilon \right)\right)=\infty \:$ 
and that's mean the series diverge.
And if $a_{n\:}$ can be positive and negative, let's look at both 
$\left(S_{n_k}\right)$ the series of the non-negative number
$\left(S_{n_r}\right)$ the series of the negative number 
that will converge to  $\infty \:,\:-\infty $
So that means ${S_n}$ doesn't have a limit
What do you say guys? Is it legit prove  ?
 A: The fundamental idea contains the germ of a correct proof and I'd give serious part credit on an exam. 
Edit: but the fix I suggested isn't right. I've left it appended so you can see that teachers make mistakes too. Thanks to @Hetebrij .
The main problem in the proposed proof by contradiction isn't that the negation of "$a_n$ converges to $0$" isn't "$a_n$ converges to something else", it's "$a_n$ doesn't converge to $0$". That means "there is some $\epsilon > 0$ such that $|a_n| > \epsilon$ infinitely often". 
But the fact that the original series converges means that eventually the partial sums are all closer together than $\epsilon /2$. That's impossible if the terms are often bigger than $\epsilon$, hence the contradiction.

Original faulty answer:
The main problem is that you don't say where $\epsilon$ comes from or what its value is in the second line, so you can't later claim that $n(L-\epsilon)$ grows without bound.
You can fix that if you remember that the definition of "limit" says (in part) "for every $\epsilon$". So you get to pick one that helps you. Try $\epsilon = L/2$.
You've also assumed $L$ is positive (thanks to @AhmedHussein for noticing). If it's not, then apply your argument to the negation of the original series.
There's no need for the later stuff dealing with the possibility that the elements of the original series might have different signs.
