Kinda' stuck,

Let (Sn) be a nondecreasing sequence of positive real numbers and define Tn= (1/n)(s1+s2+· · ·+sn)for n ∈ N. Prove that (Tn) is a nondecreasing sequence.

Thanks.

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- Sep 28th 2008, 12:17 PMhercules[SOLVED] sequence of agerages
Kinda' stuck,

Let (Sn) be a nondecreasing sequence of positive real numbers and define Tn= (1/n)(s1+s2+· · ·+sn)for n ∈ N. Prove that (Tn) is a nondecreasing sequence.

Thanks. - Sep 28th 2008, 12:33 PMhercules
Just understood it....I think I got it. Thanks

- Sep 28th 2008, 12:34 PMMoo
Hi,

Would you mind showing it please ? It can be helpful to others (Nod)

Edit : by the way, you can prove that if the sequence (Sn) converges to L, then the sequence (Tn) converges to L. This is a property of the Cesāro mean - Sep 28th 2008, 02:21 PMhercules
Hmm..I don't know about Cesaro mean but Sn in this case is only given as a non decreasing sequence...nothing else...so I don't know how I would proceed with that.

Sorry, I am not good with Latex....so here is the proof informally.

Given that Sn is a non-decreasing sequence of positive numbers, each subsequent term is greater than the previous for all 'n' in**N**.

Then I wrote down several terms for the sequence Tn:

T1 = (s1)/1

T2 = (s1+s2)/2

T3 = (s1+s2+s3)/3

and noticed that T2 - T1 was positive , T3 - T2 was positive and so on...

Since we are proving that Tn is a non-decreasing sequence as well...its every subsequent term has to be greater than the previous.

That is,

T(n+1)-Tn >=0

T(n+1)-Tn = [s1+s2+...+sn+s(n+1)]/(n+1) - [s1+s2+s3+....+sn]/n

Now to subtract make the common denominator, and get

[n*Sn+1 -(s1+s2+s3...+sn)]/ (n*(n+1))

This is greater than or equal to zero because we have n*Sn+1 and each is greater than each s1 thru. sn term.

Then I used induction to show that

T(n+1)+1 - Tn+1 is also greater than or equal to zero. Therefore the sequence must be non-decreasing.

Hope this is understandable.