Prove that a converging sequence is bounded

In summary, if a sequence {an} converges, it is also bounded. This can be proven by using the definition of convergence and the fact that for every delta>0, there exists a number N such that for every n>=N, |an - x|< delta. Therefore, for n>=N, the sequence is bounded between -delta+x and delta+x. Additionally, for n<N, an is also bounded as there are only a finite number of elements in the sequence before it starts to converge. This can be proven by induction.
  • #1
kehler
104
0

Homework Statement


Suppose that the sequence {an}converges. Show that the sequence {an} is bounded.

The Attempt at a Solution


Since the sequence converges, for every delta>0, there must exist a number N such that for every n>=N,
|an - x|< delta. Therefore, for n>=N, -delta+x < an < delta + x.
So I've proven that for n>=N, the sequence is bounded between -delta+x and delta+x.

But I don't know how to prove that for n<N, an is also bounded. I know that there are only a finite number of elements before the sequence starts to converge. Is there a theorem stating that all finite sets are bounded?
 
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  • #2
kehler said:
But I don't know how to prove that for n<N, an is also bounded. I know that there are only a finite number of elements before the sequence starts to converge. Is there a theorem stating that all finite sets are bounded?

I would consider looking at the maximum of a finite set.
 
  • #3
To expand a little:
Pick an [tex] \varepsilon > 0 [/tex], and use convergence to conclude that

[tex]
|x_i - a | < \varepsilon
[/tex]

for all [tex] i \ge N [/tex].

You now have two sets of elements of the sequence: those with [tex] i \ge N [/tex] and those for smaller [tex] i [/tex]. You should be able to argue that both sets are bounded, which means ...
 
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  • #4
Yes. All finite sets are bounded. Prove it by induction. (If you really need a proof).
 

FAQ: Prove that a converging sequence is bounded

What is a converging sequence?

A converging sequence is a sequence of numbers that approaches a specific value or limit as the number of terms in the sequence increases.

How do you prove that a converging sequence is bounded?

To prove that a converging sequence is bounded, you need to show that there exists a number M such that all terms in the sequence are less than or equal to M. This can be done by finding the limit of the sequence and showing that it is finite, or by using the definition of a converging sequence to show that the terms eventually become arbitrarily close to each other.

Why is it important to prove that a converging sequence is bounded?

Proving that a converging sequence is bounded is important because it helps us understand the behavior of the sequence and whether it will continue to approach a specific value or diverge to infinity. It also allows us to make accurate predictions and calculations based on the properties of the sequence.

What happens if a converging sequence is not bounded?

If a converging sequence is not bounded, it means that the terms in the sequence are not limited in size and can continue to increase without approaching a specific value. This is known as a diverging sequence, and it does not have a limit or a specific value that it approaches.

Can a sequence be converging and unbounded?

No, a sequence cannot be both converging and unbounded. A converging sequence, by definition, approaches a specific value or limit, which means that it is bounded. If a sequence is unbounded, it means that it does not have a limit or a specific value that it approaches, which contradicts the definition of a converging sequence.

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