Does Converging Sequence s_n Imply Converging Average t_n?

  • Thread starter Thread starter philosophking
  • Start date Start date
  • Tags Tags
    Analysis Sequences
AI Thread Summary
The discussion centers on the relationship between the convergence of a sequence s_n and its average t_n. The user questions whether the limit of t_n approaches the same value as s_n when s_n converges to a finite limit s. They express skepticism, suggesting that if s is finite, the average t_n could approach zero, which would contradict the convergence of s_n. The conversation emphasizes the need to apply the epsilon-delta definition of limits to demonstrate the relationship. Ultimately, the analysis aims to clarify how the average of a converging sequence behaves as n approaches infinity.
philosophking
Messages
175
Reaction score
0
Analysis problem (sequences)--please help

Here is the definition:

t_n = [s_1 + s_2 + ... + s_n] / n ; n >/= 1

I have to show that if lim n-> [infinity] s_n = s, then lim n-> [infinity] t_n = s

First of all, I don't think it's true. Because if s is finite, then lim s/n as n-> [infinity] would be zero, right? And thus lim t_n as n-> [infinity] is zero, and they're not the same.

I'm just wondering how to go about this problem. Thank you.
 
Physics news on Phys.org
I think I would try it directly with the epsilon-delta definition of a limit.


BTW, you should be able to convince yourself that the limit of t_n is not always zero by considering a simple example.
 


The analysis problem you have presented deals with sequences, specifically the limit of a sequence. In order to solve this problem, we need to understand the definitions and properties of limits and sequences.

A sequence is a list of numbers arranged in a specific order. In this case, the sequence is denoted by t_n and is defined as the average of the first n terms of another sequence, denoted by s_n. The average is calculated by adding all the terms and dividing by the number of terms.

A limit is the value that a sequence approaches as the number of terms increases. In this case, we are interested in the limit as n approaches infinity. This means that we are looking at the behavior of the sequence as the number of terms becomes larger and larger.

Now, let's look at the first statement: if lim n-> [infinity] s_n = s. This means that as n approaches infinity, the terms of the sequence s_n get closer and closer to the value s. In other words, the terms of the sequence s_n are approaching the limit s.

Next, we have to show that if this statement is true, then lim n-> [infinity] t_n = s. This means that if the terms of the sequence s_n are approaching the limit s, then the terms of the sequence t_n, which are the averages of the terms of s_n, are also approaching the limit s.

To prove this, we can use the definition of the limit. We have to show that for any small positive number ε, there exists a positive integer N such that for all n > N, the difference between t_n and s is less than ε. In other words, we have to show that t_n is getting closer and closer to s as n approaches infinity.

Since we know that lim n-> [infinity] s_n = s, we can choose a large enough N such that for all n > N, the difference between s_n and s is less than ε/2. This means that for all n > N, we have:

|s_n - s| < ε/2

Now, let's look at the definition of t_n:

t_n = [s_1 + s_2 + ... + s_n] / n

We can rewrite this as:

t_n = [(s_1 - s) + (s_2 - s) + ... + (s_n - s) + ns] / n

Notice that
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top