Cesaro-like summability

[Bernard]

Hi,

Suppose a(n)>0 is a periodic sequence with (a(1)+a(2)+..+a(n))/n--> M
as n-->infty.

Let b(n)>0 be a nonperiodic bounded sequence.

are there simple conditions to add to b(n) in order to have:

(a(1)b(1)+a(2)b(2)+...+a(n)b(n))/(b(1)+b(2)+....+b(n))-->M as n--
>infty.

Thanks for help.

[Robert Israel]

Bernard <benoi...@orange.fr> writes:

One simple sufficient condition is lim_{n -> infty} b(n) = c > 0.
There are more complicated conditions. For integers A < B, let
V(A,B) = max_{A <= j < B} b(j) / min_{A <= j < B} b(j). Then
a sufficient condition is that
1) lim_{n->infty} V(n,n+P) = 1, where P is the period of the
sequence {a(n)}, and
2) sum_{n=1}^infty b(n) = infinity
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

[Bernard]

On 24 f�v, 15:00, Robert Israel <isr...@math.MyUniversitysInitials.ca>
wrote:

Thank you very much,

The complicated conditions are fine but useless for me since V(n,n+P)
takes a finite number of values among more than P distinct values.

I suppose the weaker condition:

(b(1)+b(2)+...+b(n))/n-->c>0 as n-->infty is not sufficient. If so, do
you see something to add to this condition for working?

I mean something like:

b(1)+b(2)+...+b(n)=c*n+o(f(n))

for some suitable function like f(n)=log(n) or f(n)=sqrt(n)?

Thanks again.

BT

[Robert Israel]

Bernard <benoi...@orange.fr> writes:

That can't work. Consider e.g.
a(n) = b(n) = 1 if n is even, 2 if n is odd

(a(1) + ... + a(n))/n -> M = 3/2,
(a(1) b(1) + ... + a(n) b(n))/(b(1) + ... + b(n)) -> 5/3
and b(1) + ... + b(n) = 3/2 n + O(1).

[Bernard]

On 26 f�v, 23:00, Robert Israel <isr...@math.MyUniversitysInitials.ca>

> University of British Columbia � � � � � �Vancouver, BC, Canada- Masquer le texte des messages pr�c�dents -
>
> - Afficher le texte des messages pr�c�dents -

Yes of course but I still consider b to be nonperiodic with the
additional condition (b(1)+...+b(n))/n--> some limit

There is a general case for which it seems working. If a and b are 2
periodic sequences with period lenght L1 and L2 respectively and such
that gcd(L1,L2)=1.

[Robert Israel]

Bernard <benoi...@orange.fr> writes:

> Yes of course but I still consider b to be nonperiodic with the

> additional condition (b(1)+...+b(n))/n--> some limit
>
> There is a general case for which it seems working. If a and b are 2
> periodic sequences with period lenght L1 and L2 respectively and such
> that gcd(L1,L2)=1.

Yes, of course that case will work, because in any L1 L2 consecutive
integers each a(i), i=1..L1, gets matched once with each b(j), j=1..L2,
resulting in
a(1) b(1) + ... + a(n L1 L2) b(n L1 L2)
= n (a(1) + ... + a(L1))(b(1) + ... + b(L2)) = n L1 L2 M c