Let $(a_n)$ be a convergent sequence. Suppose that $\lim_{n\to\infty} a_n > 0$. Use the definition of a limit to prove that there exists N ∈ (N) (naturals) such that $a_n$ > 0 for all n ≥ N.
Use definition of a limit ($|a_n−L|<ϵ$).
Please check my proof!
My Proof:
Let $\epsilon$ be an arbitrary positive number. Let L = $\lim_{n\to\infty} a_n > 0$. Choose N ∈ (N) (naturals) such that n ≥ N implies that $|a_N−L|<\frac{\epsilon}{2}$ by definition of a limit. Assume n ≥ N. Therefore, $|2a_N−2L|=|2(a_N−L)|=|2||a_N-L|<\frac{\epsilon}{2}*2=\epsilon$. Since L > 0, $a_n$ > 0.
Best Answer
I don't think your proof is quite right. Instead, try expanding the definition of a limit a little bit. The key will be to pick some $\varepsilon >0$ that is strictly less than the limit $L$, say $\varepsilon = \frac{L}{2}$ (we can do that because $L$ is positive). Now note that there exists $N\in \mathbb{N}$ such that for all $n\geq N$ we have $$\lvert a_n - L\rvert < \varepsilon \iff - \varepsilon < a_n - L < \varepsilon.$$ Adding $L$ results in $$L-\varepsilon < a_n < L + \varepsilon.$$ Since we chose our $\varepsilon = \frac{L}{2}$, we have $$L-\varepsilon = L - \frac{L}{2} = \frac{L}{2} > 0,$$ because $L>0$. Hence $0< a_n$ for all $n\geq N$, as required.