1. ## Sequences

If a real sequence converges,then it is bounded.
How can we prove this statement?

2. Let its limit be L. The very definition of convergence implies the existence of a natural number N such that $|a_{n}-L|<1$ for every $n \geq N$. The former inequality implies in turn that $|a_{n}| < 1+|L|$ for every $n \geq N.$

Hence, starting from the N-th element of the sequence you can place them all inside of a disk of radius 1+|L| (and centered at the origin). it's clear that up to this moment some elements of the sequence might remain outside of this disk, but since it's only a finite number of them, we are done, right?

3. I think if we choose :

K = max{ $|a_{1}|,|a_{2}|.........|a_{N}|,|L| +1$} then we will have for all nεN $|a_{n}|\leq K$

4. Originally Posted by xalk
I think if we choose :

K = max{ $|a_{1}|,|a_{2}|.........|a_{N}|,|L| +1$} then we will have for all nεN $|a_{n}|\leq K$
A little clarification: N is the positive integer such that if n> N, then |a_n- L|< 1 which is guarenteed to exist since L is the limit of the sequence.

5. Originally Posted by HallsofIvy
A little clarification: N is the positive integer such that if n> N, then |a_n- L|< 1 which is guarenteed to exist since L is the limit of the sequence.
Well ,i think coquitao clarified that in his post,did he not?

6. Originally Posted by coquitao
Let its limit be L. The very definition of convergence implies the existence of a natural number N such that $|a_{n}-L|<1$ for every $n \geq N$. The former inequality implies in turn that $|a_{n}| < 1+|L|$ for every $n \geq N.$

Hence, starting from the N-th element of the sequence you can place them all inside of a disk of radius 1+|L| (and centered at the origin). it's clear that up to this moment some elements of the sequence might remain outside of this disk, but since it's only a finite number of them, we are done, right?
How do you get from $|a_{n}-L|<1$ to $|a_{n}| < 1+|L|$? You might need some explanation to that. Could use reverse triangle inequality to show that $|a_n-L|\geq |a_n|-|L|$

7. $|a_n|=|a_n-L+L|\leq|a_n-L|+|L|\implies|a_n|-|L|\leq|a_n-L|$

Using this you have $|a_n|-|L|\leq|a_n-L|<1\implies |a_n|-|L|<1\implies |a_n|<1+|L|$