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?
Best Answer
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.