Converging Sequence: Basic Steps and Practice Problems

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In summary, you need to start with an arbitrary convergent sequence and use what you know about the limits to solve the problem.
  • #1
Lee33
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Homework Statement



I was given this homework problem:

Show that if ##a_1,a_2, ... ,## is a sequence of real numbers that converges to ##a##, then [tex]lim_{n\to \infty}\frac{\sum^n_{k=1} a_k}{n}=a.[/tex]

I was provided a solution but my book never went over such examples or the concrete steps to solve such a problem. I am wondering what are the basic first steps to solving these types of problems?

And if possible, where can I find practice problems like these online? I searched but I couldn't find any.
 
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  • #2
Hint:
Remember that for every n, you may rewrite:
[tex]na=\sum_{k=1}^{n}a[/tex]
 
  • #3
Lee33 said:

Homework Statement



I was given this homework problem:

Show that if ##a_1,a_2, ... ,## is a sequence of real numbers that converges to ##a##, then [tex]lim_{n\to \infty}\frac{\sum^n_{k=1} a_k}{n}=a.[/tex]

I was provided a solution but my book never went over such examples or the concrete steps to solve such a problem. I am wondering what are the basic first steps to solving these types of problems?

There are various techniques for limit problems, but since this problem asks you to start with an arbitrary convergent sequence [itex](a_k)[/itex] the only one which will work is to use what you know about [itex](a_k)[/itex]: for all [itex]\epsilon > 0[/itex] there exists [itex]K \in \mathbb{N}[/itex] such that if [itex]k \geq K[/itex] then [itex]a - \epsilon < a_k < a + \epsilon[/itex].

That suggests taking an arbitrary [itex]\epsilon > 0[/itex] and its corresponding [itex]K[/itex] and splitting the sum as follows:
[tex]
\frac1n \sum_{k=1}^n a_k = \frac1n \sum_{k=1}^{K-1} a_k + \frac1n \sum_{k=K}^n a_k
[/tex]
(You are interested in the limit [itex]n \to \infty[/itex], so at some stage you will have [itex]n > K[/itex] and you may as well assume that to start with.)

Your plan is to show that
[tex]
a - \epsilon \leq \lim_{n \to \infty} \frac1n \sum_{k=1}^n a_k \leq a + \epsilon
[/tex]
and since [itex]\epsilon > 0[/itex] was arbitrary it must follow that
[tex]
\lim_{n \to \infty} \frac1n \sum_{k=1}^n a_k = a
[/tex]
as required.
 
  • #4
I would use ϵ/2 in one step instead of ϵ, that makes the inequalities easier to show.
 
  • #5
Thank you very much, pasmith! That cleared some issues I had, thanks for clarifying it for me.
 

What is a sequence that converges?

A sequence that converges is a mathematical concept where the terms of a sequence approach a specific value as the sequence progresses. This value is known as the limit of the sequence.

How do you know if a sequence converges?

A sequence converges if the terms of the sequence get closer and closer to the limit as the sequence progresses. This can be determined by calculating the difference between consecutive terms and seeing if it approaches zero as the sequence continues.

What is the limit of a convergent sequence?

The limit of a convergent sequence is the value that the terms of the sequence approach as the sequence progresses. It is the value that the sequence converges to.

How do you prove that a sequence converges?

To prove that a sequence converges, you must show that the terms of the sequence get closer and closer to the limit as the sequence progresses. This can be done algebraically or using techniques such as the squeeze theorem or the monotone convergence theorem.

What is the difference between a convergent sequence and a divergent sequence?

A convergent sequence approaches a specific limit as the sequence progresses, while a divergent sequence does not have a specific limit and the terms of the sequence either increase or decrease without approaching a fixed value.

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