Expert: Paul Klarreich - 4/29/2009

Question
Hi Paul, first, thanksfor prior help, has been great.

I have another question on convergence, which I need a start on...

Let an arbitrary convergent sequence {a_n} have limit A.

Show the sequence {b_n} given by {(a_1 + a_2 _...+ a_n)/n} also converges to A.

Hints: Suppose a_n -> A and use the tolerance e/2 to ensure that |a_n - A| < e/2 for all sufficiently large n, n > N for some given N.
Choose M sufficiently large so that for n > M > N, |b_n - A| = |(a_1 +...+a_N - NA)/n + (a_(n+1) +...+a_n - (n - N)A)/n| is less than e.  

Answer
Questioner:   Steve
Country:  Australia
Category:  Advanced Math
Private:  No
 
Subject:  Advanced Maths
Question:  Hi Paul, first, thanksfor prior help, has been great.

I have another question on convergence, which I need a start on...

Let an arbitrary convergent sequence {a_n} have limit A.

Show the sequence {b_n} given by {(a_1 + a_2 _...+ a_n)/n} also converges to A.

Hints: Suppose a_n -> A and use the tolerance e/2 to ensure that |a_n - A| < e/2 for all sufficiently large n, n > N for some given N.
Choose M sufficiently large so that for n > M > N, |b_n - A| = |(a_1 +...+a_N - NA)/n + (a_(n+1) +...+a_n - (n - N)A)/n| is less than e.
....................................................
Hi, Steve,
[I have been very busy this week and this one took some work.]

So lim a[n] = A

And b[n] = 1/n SUM( a[k] )

Now suppose we have N such that for all k > N, | a[k] - A | < e/2

What is  | b[n] - A | ?
   a[1] + ... + a[n] - nA
= |------------------------ |  
          n

  (a[1] - A)  + ... + (a[n] - A)
= |------------------------------ |
          n

  |a[1] - A|  + ... + |a[n] - A|
< |------------------------------ |
               n

Now lets call those | a[k] - A | things deviations:

 d[1]  + d[2] ... + d[n]
-----------------------
             n

Now this n is bigger than the N. (Sounds strange that little n is larger than big N, but little n increases while big N is fixed by that e/2 thing.)

So we split the sum into two parts: before N and after N.


  d[1]  + d[2] + ... d[N]     d[n+1] ... + d[n]
|------------------------- + -------------------- |
              n                       n

Now the sum  d[1]  + d[2] + ... d[N] is FIXED, since N is fixed.  So we can call that Big-D. (as in Big D, little-a, double-l, a, s spells ... oh, never mind.)


Therefore the first fraction depends only on the denominator and that can be made big,(that's your Big-M) so Big-D/n can be made small, in fact we can take n big enough that Big-D/n < e/2.  [hence that hint you got.]

Now what about the second fraction?

  d[n+1] ... + d[n]
 --------------------
          n

Now each of THOSE d[]'s < e/2.
There are fewer than n of them.
So their sum is < ne/2
But the sum is divided by n.
So the fraction is < e/2.

I think that does it.

BAD NEWS -- I don't remember a lot about uniformly continuous functions, so the next question might not get an answer.  But keep them coming -- I'm having fun even if I have trouble nailing them.