Prove that a convergent sequence with non-negative terms converges to a non-negative limit Assume $a_n$ is a non-negative sequence, that is, $a_n \geq 0$ for every $n \in \mathbb{N}$. Prove that if $a_n$ converges then the limit is non-negative. Clue: prove it by negation.
I think proof by contradiction is a good method, which is $a_n$ converges and the limit is negative. Since $a_n$ converges, $\lim_{n\to\infty}a_n = L$. Also, since $a_n$ is non-negative sequence, $L \geq 0$. This is a contradiction. Is my method correct? 
 A: You have assumed what you are asked to prove.
If possible let $a_n \to L$ with $L <0$. Take $\epsilon =-L$. Then there exits $n_0$ such that $|a_n-L | <\epsilon =-L$ for $n >n_0$. But then $a_n-L \leq |a_n-L|<(-L)$, so $a_n <0$ for $n>n_0$ which is a contradiction. 
A: You seemed to have used the result you are trying to prove in your proof! This is not allowed in mathematics. 
The clue suggesting contradiction seems to be a good approach. Indeed, suppose $a_n \longrightarrow L$ where $L < 0$. Then, there is an $\epsilon > 0$ such that $-\epsilon = L$ (just take $\epsilon = -L$). Then, since the sequence converges, we are guaranteed a natural number $N$ such that $n \geq N$ implies $$|a_n - L| < \epsilon$$
Unpacking the definition of the absolute value gives us:
$$-\epsilon < a_n - L < \epsilon $$
Now recall that $-L = \epsilon$, which gives:
$$-\epsilon < a_n + \epsilon < \epsilon$$
Subtracting $\epsilon$ from both sides of the above inequality yields:
$$-2\epsilon < a_n < 0$$
This contradicts the fact that $a_n \geq 0$ for all $n \in \mathbb{N}$, so we are done.
A: One way to do this is by contrapositive: if $L \in \mathbb{R}$ is negative, show that the sequence does not converge to $L$. Take $L \in \mathbb{R}$ with $s<0$. Then you know that for every $n$ you must have $|a_n - L| \geq |s|$ (why?). 
Now, use the definition of the limit of a sequence to show that the sequence cannot converge to $L$. 
