# Eventually bounded sequence is bounded vice-versa

A sequence {$$x_n$$} is said to be eventually bounded if there exists a positive integer $$N$$ and a positive number $$M$$ such that |$$x_n$$|< M for all $$n>N$$. Prove that a sequence is bounded if and only if it is eventually bounded.

Progress: Since this is an iff statement, I was able to prove the first one.

(→) Suppose sequence {$$x_n$$} is eventually bounded. We will show that {$$x_n$$} is bounded. Since after $$N$$ terms the sequence is bounded, we will consider the finite set {$$x_1,x_2,…,x_N$$} and $$x_{N+i}$$ for $$i\in N$$ (which we know is bounded). Let $$U_1$$ and $$L_1$$ be the upper and lower bound of {$$x_1,x_2,…,x_N$$} respectively and $$U_2$$ and $$L_2$$ be the upper and lower bound of $$x_{N+i}$$ for $$i\in N$$ respectively. Moreover, let $$U$$ be the upper bound of {$$x_n$$} then $$U=max⁡(U_1,U_2)$$ and similarly, let $$L$$ be the lower bound of {$$x_n$$} then $$L=max⁡(L_1,L_2)$$. So, {$$x_n$$ } has both upper and lower bound then {$$x_n$$} is bounded.

(←)Suppose sequence {$$x_n$$} is bounded. We will show that {$$x_n$$} is eventually bounded. "I am stuck here. Please help! Any hint or help will be much appreciated."

• Just take $N=1$. – TonyK Apr 7 at 15:24
• Well, if it is bounded, you can take any bound for $M$, and any value for $N$. He who can do the more can do the least. – Bernard Apr 7 at 15:25
• What do you mean? Can you clarify further? – mathhunterx Apr 7 at 15:25
• If the sequence is bounded, it is a fortiori eventually bounded. – Bernard Apr 7 at 15:28
• So you mean no need for formal proof? – mathhunterx Apr 7 at 15:37