Convergence of sequences

1. Sep 14, 2010

swuster

1. The problem statement, all variables and given/known data
If {cn} is a convergent sequence of real numbers, does there necessarily exist R> 0 such that |cn|≤ R for every n ∈ N? Equivalently, is {cn : n ∈ N} a bounded set of real numbers? Explain why or why not.

2. Relevant equations
n/a

3. The attempt at a solution
I would think this is patently obvious given the definition of convergence but I don't really know how to put the proof in words and numbers.

cn approaches some value c for large n but it can do so in a number of ways so how can i prove that there is always some R that is larger or equal to all elements cn?

Thanks for the help!

2. Sep 14, 2010

JonF

Write the limit definition of a sequence out. Assume to opposite of your claim (for contradiction).
So for any N you pick to satisfy an epsilon of the limit, there will be an R as large as you want where R < |Xn|. Where epsilon is fixed already. So your R can be dependant on epsilon and L. Is there some R that would bring about a contradiction?

3. Sep 14, 2010

HallsofIvy

Suppose your sequence converges to L. Then, by definition of "limit", given any $\epsilon> 0$ there must exist some N such that if n> N then $|a_n- L|< \epsilon$.

Take $\epsilon= 1$, say. Then there exist N such that if n> N, $|a_n- L|< 1$ which is the same as saying $-1< a_n- L< 1$ or $L-1< a_n< L+ 1$. Now that only restricts $x_n$ for n> N, but the set $\{a_0, a_1, a_2, \cdot\cdot\cdot, a_N\}$ is finite and has a largest member. Every number in the sequence $\{a_n\}$ must be less that the that largest member or L+ 1, whichever is larger.

4. Sep 14, 2010

swuster

In my case {cn} need not be finite, though. Does that still hold in this case?

5. Sep 14, 2010

Staff: Mentor

Yes. HallsOfIvy is talking about the terms in the sequence c1, c2, ..., cN, the first N terms at the beginning of the sequence.