Prove that the limit of a sequence exists

[shrub_broom]

Homework Statement

suppose that 0≤x_m+n≤x_m+x_n for all m,n∈ℕ, prove that the limit of x_n/n exists when n tends to infinity.

Homework Equations

The Attempt at a Solution

I get that x_n is bounded by zero and x₁. And I guess that x_n is monotonous but i find it hard to prove. Or maybe there is another way to prove the existence of this limit.

 

[LCKurtz]

Homework Statement

suppose that 0≤x_m+n≤x_m+x_n for all m,n∈ℕ, prove that the limit of x_n/n exists when n tends to infinity.

Homework Equations

The Attempt at a Solution

I get that x_n is bounded by zero and x₁. And I guess that x_n is monotonous but i find it hard to prove. Or maybe there is another way to prove the existence of this limit.

Perhaps it is just a typo, but $x_n$ is not bounded above by $x_1$. It isn't bounded above at all. Think about the sequence $x_n = n$. It satisfies all the conditions and you have $\frac {x_n} n \to 1$. That seems like a worst case sequence, but I don't have anything further for you at the moment. Not too helpful I know...

 

[shrub_broom]

Well, since for each m∈ℕ, we get -

Perhaps it is just a typo, but $x_n$ is not bounded above by $x_1$. It isn't bounded above at all. Think about the sequence $x_n = n$. It satisfies all the conditions and you have $\frac {x_n} n \to 1$. That seems like a worst case sequence, but I don't have anything further for you at the moment. Not too helpful I know...

well i just made a mistake. and i want to mean x_n/n is bounded by x₁. sorry for that fault

 

[shrub_broom]

the sequence is not monotonous. Let x_2n=a and x_2n+1=b; a≤b≤2a.

 

[LCKurtz]

the sequence is not monotonous. Let x_2n=a and x_2n+1=b; a≤b≤2a.

You are correct, the sequence is not monotonous (boring). It is also not monotone.

 

[Buzz Bloom]

the sequence is not monotonous. Let x2n=a and x2n+1=b; a≤b≤2a

You are correct, the sequence is not monotonous (boring). It is also not monotone.

Hi shrub and LC:
I am confused by the conclusion that the series is not monotonic. What is the rational for this conclusion from the statement

(1) x_2n=a and x_2n+1=b; a≤b≤2a?​

Even when adding the PO requirement

(2) 0≤x_m+n≤x_m+x_n​

I do not see this conclusion as valid.

Suppose we define

x_n = n.​

This is a monotonic sequence that satisfied both (1) and (2).
Using (1)

2n ≤ 2n+1 ≤ 4n.​

Using (2)

m + n ≤ m + n.​

Regards,
Buzz

 

[Delta2]

From the given inequality for $m=n+1$ we have that $$x_{n+1}\leq x_n+x_1 \Rightarrow \frac{x_{n+1}}{n+1}\leq \frac{x_n+x_1}{n+1}\leq \frac{x_n}{n}+\frac{x_1}{n}$$, so for the sequence $y_n=\frac{x_n}{n}$ we have that $$y_{n+1}\leq y_n+\frac{x_1}{n}$$.

Now this is equivalent to say that "for any $\epsilon>0$ there exists $n_0\geq \frac{x_1}{\epsilon}$ such that for $n\geq n_0$, $y_{n+1}\leq y_n+\epsilon$", so $y_n$ is not exactly monotonous it is something like monotonous in a neighbourhood of +infinity

Hope that helps.

 

[shrub_broom]

i get a proof. suppose that the limit inferior of the sequence is a, and then prove that the limit superior is no greater than a. to prove so, for any positive ε, there exist N x_N/N ≤ a+ε. and for any n∈ℕ∧n≥N, we have n =p N+q p,q∈ℕ and q is less than N. hence, x_n=x_pN+q≤px_N+N x₁; x_n/n≤a+ε+Nx₁/n. when n tends to infinity, since N, x₁ are limited, we get that the limit superior is no greater than its limit inferior. hence the sequence is converge.

 

[shrub_broom]

Hi shrub and LC:
I am confused by the conclusion that the series is not monotonic. What is the rational for this conclusion from the statement

(1) x_2n=a and x_2n+1=b; a≤b≤2a?​

Even when adding the PO requirement

(2) 0≤x_m+n≤x_m+x_n​

I do not see this conclusion as valid.

Suppose we define

x_n = n.​

This is a monotonic sequence that satisfied both (1) and (2).
Using (1)

2n ≤ 2n+1 ≤ 4n.​

Using (2)

m + n ≤ m + n.​

Regards,
Buzz

for any m,n. x_m+n≤x_m+x_n since a≤2b. if divide the right side,it becomes greater and the inequality is still true.

 

[Delta2]

i get a proof. suppose that the limit inferior of the sequence is a, and then prove that the limit superior is no greater than a. to prove so, for any positive ε, there exist N x_N/N ≤ a+ε. and for any n∈ℕ∧n≥N, we have n =p N+q p,q∈ℕ and q is less than N. hence, x_n=x_pN+q≤px_N+N x₁; x_n/n≤a+ε+Nx₁/n. when n tends to infinity, since N, x₁ are limited, we get that the limit superior is no greater than its limit inferior. hence the sequence is converge.

what happens to $p$? Shouldn't it be $\frac{x_n}{n}\leq p(a+\epsilon)+\frac{Nx_1}{n}$ but as n tends to infinity, p tends to infinity also...

 

[shrub_broom]

what happens to $p$? Shouldn't it be $\frac{x_n}{n}\leq p(a+\epsilon)+\frac{Nx_1}{n}$ but as n tends to infinity, p tends to infinity also...

i think it should be $\frac{x_n}{n}\leq\frac{px_N+qx_1}{pN+q}\leq \frac{x_N}{N} +\frac{Nx_1}{n}$

 

[Delta2]

i think it should be $\frac{x_n}{n}\leq\frac{px_N+qx_1}{pN+q}\leq \frac{x_N}{N} +\frac{Nx_1}{n}$

Ok fine , one last question from me, $N$ is not exactly a constant (like $x_1$) but it depends on $\epsilon$. It might be the case that as $\epsilon \rightarrow 0$ that $N\rightarrow \infty$ and this leads to a problem (cause we will have as $n\rightarrow \infty, \frac{Nx_1}{n}\rightarrow \frac{\infty}{\infty}$).

I believe in order to complete the proof, we should prove that $N=N(\epsilon)\leq c$ for some constant $c$.

 

[shrub_broom]

Ok fine , one last question from me, $N$ is not exactly a constant (like $x_1$) but it depends on $\epsilon$. It might be the case that as $\epsilon \rightarrow 0$ that $N\rightarrow \infty$ and this leads to a problem (cause we will have as $n\rightarrow \infty, \frac{Nx_1}{n}\rightarrow \frac{\infty}{\infty}$).

I believe in order to complete the proof, we should prove that $N=N(\epsilon)\leq c$ for some constant $c$.

Exactly i think there is no need to prove this hypothesi.Write the right of the inequality in the form of defination we get $a+\eplison+frac{Nx_1}{n}$. Although N is determined by$\eplison$, if we choose a big enough $p$, then$\frac{Nx_1}{n}$can be any positive real number, which means $\eplison+frac{Nx_1}{n}$ can also be any positive real number. And this is the defination of the Iimit.

 

[shrub_broom]

Ok fine , one last question from me, $N$ is not exactly a constant (like $x_1$) but it depends on $\epsilon$. It might be the case that as $\epsilon \rightarrow 0$ that $N\rightarrow \infty$ and this leads to a problem (cause we will have as $n\rightarrow \infty, \frac{Nx_1}{n}\rightarrow \frac{\infty}{\infty}$).

I believe in order to complete the proof, we should prove that $N=N(\epsilon)\leq c$ for some constant $c$.

Exactly i think there is no need to prove this hypothesi.Write the right of the inequality in the form of defination we get $a+\epsilon+\frac{Nx_1}{n}$. Although N is determined by$\epsilon$, if we choose a big enough $p$, then$\frac{Nx_1}{n}$can be any positive real number, which means $\epsilon+frac{Nx_1}{n}$ can also be any positive real number. And this is the defination of the Iimit.

 

[shrub_broom]

Exactly i think there is no need to prove this hypothesi.Write the right of the inequality in the form of defination we get $a+\epsilon+\frac{Nx_1}{n}$. Although N is determined by$\epsilon$, if we choose a big enough $p$, then$\frac{Nx_1}{n}$can be any positive real number, which means $\epsilon+\frac{Nx_1}{n}$ can also be any positive real number. And this is the defination of the Iimit.

Well, I still forget to correct some typo.

 

[Delta2]

I think you are right after all. We can choose $n$ (or $p$ as you say) as big as we want (for example $n=N^2$), and thus $\frac{Nx_1}{n}$ can become as small as we want.

 

[Buzz Bloom]

for any m,n. xm+n≤xm+xn since a≤2b. if divide the right side,it becomes greater and the inequality is still true.

Hi shrub:

I think I now understand my confusion, but since I am not sure, I would much appreciate your response.

There are two being discussed: x_n and x_n/n.

In my post #6, I thought your statement in post #4

the sequence is not monotonous

referred to the sequence x_n in a general way, and I believe this sequence may or may not be monotonic. However,I now think you meant that the sequence x_n/n is not monotonic. Is this correct?

Regards,
Buzz

 

[shrub_broom]

Hi shrub:

I think I now understand my confusion, but since I am not sure, I would much appreciate your response.

There are two being discussed: x_n and x_n/n.

In my post #6, I thought your statement in post #4

referred to the sequence x_n in a general way, and I believe this sequence may or may not be monotonic. However,I now think you meant that the sequence x_n/n is not monotonic. Is this correct?

Regards,
Buzz

yes, the x_n/n is not monotonous