Answer to Question #161248 in Calculus for Vishal

Question

Let {an}∞ n=1 be a convergent sequence with limit L such that B≤an forevery n≥N for some N∈ N and some B∈R. Prove that B≤L

Answer 1

Let us prove by contraposition. Suppose that $B>L$. Therefore, $B-L>0$. Let $\epsilon =\frac{B-L}{2}$. Since $\{a_n\}_{n=1}^{\infty}$ be a convergent sequence with limit $L$, there exist $M\in\mathbb N$ such that $|a_n-L|<\epsilon$ for every $n\ge M$. Let $k\ge \max\{N, M\}$. Then $-\epsilon< a_k-L<\epsilon$. It follows that $a_k<\epsilon + L=\frac{B-L}{2}+L=\frac{B+L}{2}<\frac{B+B}{2}=B$. Consequently, $a_k<B$, and we have a contradiction with $a_k\ge B$. Therefore, $B\le L.$