Is Every Convergent Sequence of Real Numbers Bounded?

[swuster]

Homework Statement

If {c_n} is a convergent sequence of real numbers, does there necessarily exist R> 0 such that |c_n|≤ R for every n ∈ N? Equivalently, is {c_n : n ∈ N} a bounded set of real numbers? Explain why or why not.

Homework Equations

n/a

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.

c_n 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 c_n?

Thanks for the help!

 

[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?

 

[HallsofIvy]

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

Take \epsilon= 1, say. Then there exist N such that if n> N, |a_n- L|&lt; 1 which is the same as saying -1&lt; a_n- L&lt; 1 or L-1&lt; a_n&lt; 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.

 

[swuster]

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

 

[Mark44]

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

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