For direct proof, how do you choose M for bounded sequence?

[mikeyBoy83]

So the definition of a bounded sequence is this:
A sequence $(x_{n})$ of real numbers is bounded if there exists a real number $M>0$ such that $|x_{n}|\le M$ for each $n$

My question is pretty simple. How does one choose the M, based on the sequence in order to arrive at the conclusion? This has been something I've been confused about for a number of years and it seems like it all depends on the sequence in question. But is there a general rule that applies in these cases?

 

[PeroK]

It's difficult to see what is the confusion. Do you have an example in mind? Clearly no single $M$ will work for all bounded sequences.

 

[mikeyBoy83]

It's difficult to see what is the confusion. Do you have an example in mind? Clearly no single $M$ will work for all bounded sequences.

Right I understand no single $M$ works for all bounded sequences. But specifically let's consider something like $a_n=\frac{sin(n)}{n}$, for example. We basically need to choose $M$ such that $|\frac{sin(n)}{n}|\le M$ for each $n$. The question is how do I go about choosing $M$ in a case like this? Thanks!

 

[PeroK]

Right I understand no single $M$ works for all bounded sequences. But specifically let's consider something like $a_n=\frac{sin(n)}{n}$, for example. We basically need to choose $M$ such that $|\frac{sin(n)}{n}|\le M$ for each $n$. The question is how do I go about choosing $M$ in a case like this? Thanks!

You need to understand the sequence. Why is it bounded?

 

[mikeyBoy83]

You need to understand the sequence. Why is it bounded?

No matter what $n$ we choose, the distance between consecutive terms of $a_{n}$ are within some finite range. Basically, it converges, but how does that dictate what my proof involving the definition should be written?

Edit: I just thought of this. Suppose we choose $M\ge 1/n$, then for each n

$|\frac{sin(n)}{n}|\le |\frac{1}{n}|\le |M|$

And since $M>0$, $|a_{n}|\le M$. Hence, $a_{n}$ is bounded.

 

[FactChecker]

For a particular sequence, the existence of any M like that will make it bounded. But to prove it is bounded, you need to look at the sequence and find a specific M for your proof. The proof is designed around your selection of M.

 

[PeroK]

No matter what $n$ we choose, the distance between consecutive terms of $a_{n}$ are within some finite range. Basically, it converges, but how does that dictate what my proof involving the definition should be written?

Edit: I just thought of this. Suppose we choose $M\ge 1/n$, then for each n

$|\frac{sin(n)}{n}|\le |\frac{1}{n}|\le |M|$

And since $M>0$, $|a_{n}|\le M$. Hence, $a_{n}$ is bounded.

Perhaps you haven't undertstood that for each sequence you need to find a specific number. In your example:

For $n \ge 1$ we have $|a_n| = |\frac{\sin (n)}{n}| \le |\sin (n)| \le 1$

Hence the sequence $(a_n)$ is bounded by $1$ (and, of course by any number greater than $1$).

You could say it is bounded by $M = 1$, but in a specific case there is no need to mention $M$.

 

[mikeyBoy83]

Perhaps you haven't undertstood that for each sequence you need to find a specific number. In your example:

For $n \ge 1$ we have $|a_n| = |\frac{\sin (n)}{n}| \le |\sin (n)| \le 1$

Hence the sequence $(a_n)$ is bounded by $1$ (and, of course by any number greater than $1$).

You could say it is bounded by $M = 1$, but in a specific case there is no need to mention $M$.

Don't you mean

For $n \ge 1$ we have $|a_n| = |\frac{\sin (n)}{n}| \le |\frac{1}{n}| \le 1$ ?

I suppose it wouldn't matter since sin(n) is bounded by 1 as well.

Actually after posting this my thoughts directed me to choosing 1, which when plotted is equally justified (not a convincing argument though). Thanks for your help.

 

[mikeyBoy83]

Here is a similar example. I want to show that $a_{n} = \frac{sin(n)\cdot ln(2n)}{n}$ is bounded under the assumption that $ln(x)\le x$ for each $x \in R$.

For that choose $M=1$ as well. Then, for $n\ge 1$ we have $|a_n| = |\frac{\sin (n)\cdot ln(2n)}{n}| \le |\frac{ln(2n)}{n}|$. Since, $ln(x)\le x$ for each $x \in R$, then $ln(2n) \le n$ for any $n \in N\subset R$. Hence,

$|\frac{ln(2n)}{n}| \le |\frac{n}{n}|\le 1$

So, $a_{n}$ is bounded by $M=1$.

 

[PeroK]

Here is a similar example. I want to show that $a_{n} = \frac{sin(n)\cdot ln(2n)}{n}$ is bounded under the assumption that $ln(x)\le x$ for each $x \in R$.

For that choose $M=1$ as well. Then, for $n\ge 1$ we have $|a_n| = |\frac{\sin (n)\cdot ln(2n)}{n}| \le |\frac{ln(2n)}{n}|$. Since, $ln(x)\le x$ for each $x \in R$, then $ln(2n) \le n$ for any $n \in N\subset R$. Hence,

$|\frac{ln(2n)}{n}| \le |\frac{n}{n}|\le 1$

So, $a_{n}$ is bounded by $M=1$.

You've used $ln(2n) \le n$, which is more than you were given. You need to sort that out.

Otherwise, that's the right idea.

Any more problems need to go in Homework, though!

 

[mikeyBoy83]

You've used $ln(2n) \le n$, which is more than you were given. You need to sort that out.

Otherwise, that's the right idea.

Any more problems need to go in Homework, though!

But this isn't homework, its independent study, or does that matter?

 

[PeroK]

But this isn't homework, its independent study, or does that matter?

It doesn't matter how you're studying. For problems, you get a better response in the Homework section anyway.

 

[mikeyBoy83]

It doesn't matter how you're studying. For problems, you get a better response in the Homework section anyway.

Great! Thanks for all your help!