• Support PF! Buy your school textbooks, materials and every day products Here!

Sequence that converges

  • Thread starter Lee33
  • Start date
  • #1
160
0

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.
 
Last edited:

Answers and Replies

  • #2
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
132
Hint:
Remember that for every n, you may rewrite:
[tex]na=\sum_{k=1}^{n}a[/tex]
 
  • #3
pasmith
Homework Helper
1,740
412

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
34,281
10,322
I would use ϵ/2 in one step instead of ϵ, that makes the inequalities easier to show.
 
  • #5
160
0
Thank you very much, pasmith! That cleared some issues I had, thanks for clarifying it for me.
 

Related Threads on Sequence that converges

  • Last Post
Replies
13
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
16
Views
9K
  • Last Post
Replies
9
Views
538
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
Replies
4
Views
405
Replies
23
Views
3K
Replies
5
Views
692
Top