Prove that the limit of a sequence exists

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Homework Statement


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

Homework Equations

The Attempt at a Solution


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

Homework Statement


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

Homework Equations

The Attempt at a Solution


I get that xn is bounded by zero and x1. And I guess that xn 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...
 
Well, since for each m∈ℕ, we get -
LCKurtz said:
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 xn/n is bounded by x1. sorry for that fault
 
the sequence is not monotonous. Let x2n=a and x2n+1=b; a≤b≤2a.
 
shrub_broom said:
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. :oldsmile:
 
shrub_broom said:
the sequence is not monotonous. Let x2n=a and x2n+1=b; a≤b≤2a
LCKurtz said:
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) x2n=a and x2n+1=b; a≤b≤2a?​
Even when adding the PO requirement
(2) 0≤xm+n≤xm+xn
I do not see this conclusion as valid.

Suppose we define
xn = 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
 
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.
 
Last edited:
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 xN/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, xn=xpN+q≤pxN+N x1; xn/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.
 
Buzz Bloom said:
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) x2n=a and x2n+1=b; a≤b≤2a?​
Even when adding the PO requirement
(2) 0≤xm+n≤xm+xn
I do not see this conclusion as valid.

Suppose we define
xn = 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. xm+n≤xm+xn since a≤2b. if divide the right side,it becomes greater and the inequality is still true.
 
  • #10
shrub_broom said:
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 xN/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, xn=xpN+q≤pxN+N x1; xn/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...
 
  • #11
Delta2 said:
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}##
 
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  • #12
shrub_broom said:
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##.
 
  • #13
Delta2 said:
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.
 
  • #14
Delta2 said:
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.
 
  • #15
shrub_broom said:
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.
 
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  • #16
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.
 
  • #17
shrub_broom said:
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: xn and xn/n.

In my post #6, I thought your statement in post #4
shrub_broom said:
the sequence is not monotonous
referred to the sequence xn in a general way, and I believe this sequence may or may not be monotonic. However,I now think you meant that the sequence xn/n is not monotonic. Is this correct?

Regards,
Buzz
 
  • #18
Buzz Bloom said:
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: xn and xn/n.

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

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

Regards,
Buzz
yes, the xn/n is not monotonous
 
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