Cesaro Sum. Understanding the sequence.

[fatcrispy]

Homework Statement

I have a finite sequence Z=(z₁,...,z_n). The Cesaro sum of Z is \frac{(B₁+B₂+...+B_n)}{n}

B_C=z₁+z₂+...z_C (1\leqC\leqn)

Lets say the problem asks "The Cesaro sum of the 99th term sequence of (z₁,...,z₉₉) is 2000, what is the Cesaro sum of the 100 term sequence (1, z₁,...,z₉₉)?

Homework Equations

The Attempt at a Solution

I read about Cesaro sum on wikipedia but it didn't elaborate much. Here is where I'm at:

2000=\frac{(B₁+B₂+...+B₉₉)}{99}

But, honestly, I have no idea how to solve this because I can't find any info on it.

 

[LCKurtz]

Call z₁ + z₂ +...+z₉₉ = S

You are given that S/99 = 2000.

Now you are asked to calculate (1 + S)/100. It isn't that tough...

 

[fatcrispy]

Call z₁ + z₂ +...+z₉₉ = S

You are given that S/99 = 2000.

Now you are asked to calculate (1 + S)/100. It isn't that tough...

So, putting 1 in front of the sequence allows z₉₉ to become the 100th term? But it isn't the 100th nth term right? With what you said it would just be 1980.01 as the answer? I am trying to fundamentally understand this. I understand it's just an average but when they start saying the value of C could be lesser or equal to n and all of that I lose the concept.

 

[LCKurtz]

So, putting 1 in front of the sequence allows z₉₉ to become the 100th term? But it isn't the 100th nth term right? With what you said it would just be 1980.01 as the answer?

Yes, that is the correct answer. Add a 1 to the sequence means there are now 101 terms to average.

I am trying to fundamentally understand this. I understand it's just an average but when they start saying the value of C could be lesser or equal to n and all of that I lose the concept.

The Cesaro sum of a sequence gives you a new sequence which gives cumulative average of the given sequence. One place they are used is in the study of divergent sequences. For example, the sequence 1, -1, 1, -1, 1,... diverges. But, informally, you might say its "average value" is 0. And that is exactly what the Cesaro sum sequence converges to.

 

[fatcrispy]

Yes, that is the correct answer. Add a 1 to the sequence means there are now 101 terms to average.

Big "oh..." moment. I think I misread the problem. It asks "what is the Cesaro sum of the 100th term sequence (1, z₁, ..., z₉₉)?" I realize now that it is the whole 100 term sequence and I don't actually have to go up to z₁₀₀ right? Because the 1 in front makes it 100 terms.

The Cesaro sum of a sequence gives you a new sequence which gives cumulative average of the given sequence. One place they are used is in the study of divergent sequences. For example, the sequence 1, -1, 1, -1, 1,... diverges. But, informally, you might say its "average value" is 0. And that is exactly what the Cesaro sum sequence converges to.

I understand this. It's just the technical definition that got me. Thanks!

 

[LCKurtz]

Yes. It's 100 terms including the 1. I mistyped 101 earlier I see. I think you've got it now.