Answer to Question #162717 in Calculus for Bholu

Question

Let {an} ∞n=1 be a non-decreasing (resp. non-increasing) sequence which converges to a. Then prove that an ≤ a (resp. a ≤ an) for every n ∈ N.

Answer 1

Let $\{a_n\}_{n=1}^{\infty}$ be a non-decreasing sequence which converges to $a$. Then prove that $a_n ≤ a$ for every $n \in\mathbb N.$

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

Let $\{a_n\}_{n=1}^{\infty}$ be a non-increasing sequence which converges to $a$. Then prove that $a ≤ a_n$ for every $n \in\mathbb N.$

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