it's a fact that every convergent sequence is bounded. recall that, by definition, a sequence {An}has the limit L if for every W>0 there is a corresponding integer N such that
absolute value of (An-L)<W whereever n>N
Use this definition to prove that every sequence that converges to 0 is bounded, i.e to prove that if lim(n is infinity) An =0 then there is a positive number M such that absolute value of (An)<= M for all n belong to natural numbers.
Why not prove the fact first, and then by an obvious corollary, your statement will be true?
Proposition 3.1.4: Convergent Sequences are Bounded
Let be a sequence which converges to 0. Given any epsilon ball about 0, say you know that there exists such that for all . So simply take
and you can be sure that for all n, i.e. it is bounded.
Shannon: While well intentioned, Gamma did you no favor by posting a solution to your problem (no offense intended Gamma but I hope you will think about this too; more hints, less solution will teach more). This most fundamental question and the technique for proving it touches on about every area of mathematics. We've taken away your chance to discover the solution with the hints that you've been given as you've progressed to this point over the years.
All is not lost. You should write your own proof for a simple case on the real line in about a month without looking at Gamma's proof to make sure you know it and aren't just repeating Gamma's work. I realize that the idea isn't obvious to you now but believe me, it will pay off in your future.