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

Suppose first $a_n$ to be non-decreasing (the other case is totally symmetric). Suppose the contrary of the announced proposition, $\exists a_{N_0}, a_{N_0} >a$. As $a_n$ is non-decreasing $\forall n\geq N_0, a_n >0$. Now by taking $\epsilon = |a-a_{N_0}|>0$ we have by definition of convergence $N\in \mathbb{N}$ such that $\forall n \geq N, |a_n-a|<\epsilon$ . Now by taking $N'=\max (N_0, N)$, we get $\forall n\geq N'$ both properties : $a_n \geq a_{N_0} = a + \epsilon$, $|a_n-a|<\epsilon$. But this is clearly a contradiction, as the first inequality gives us $a_n-a \geq \epsilon$. Therefore by contradiction $\forall n, a_n \leq a$. The case of non-increasing sequence can be studied by considering the sequence $-a_n$ that is non-decreasing and converges to $-a$.