Limit proof on monotonic sequences

[Bipolarity]

Consider the sequence $\{ a_{n} \}$

If $$|a_{n+1}| > |a_{n}|$$

Prove that

$$\lim_{n→∞} a_{n} ≠ 0$$

The problem is part of a proof I am trying to understand, but I don't understand this particular step in the proof. Any ideas on how I might grasp this step?

BiP

 

[micromass]

Do you see it intuitively?? Did you draw a real number line and draw what such a sequence looks like??

Of course I can just give you the proof, but I prefer that you "see it". If you do, then I believe the proof should be very easy.

 

[Bipolarity]

Do you see it intuitively?? Did you draw a real number line and draw what such a sequence looks like??

Of course I can just give you the proof, but I prefer that you "see it". If you do, then I believe the proof should be very easy.

So because the sequence increases forever, it can't be 0 at infinity because if it were, then the term after that would be positive contradicting the fact that 0 is the limit?

But how would I prove this using the ε-δ definition?

BiP

 

[micromass]

Yeah, so basically, the closest the sequences ever gets to 0 is in $|a_0|$. After that it gets further and further away.

So, what is the definition of $\lim_{n\rightarrow +\infty}{a_n}=0$?? Can you find an ε that works?

 

[Bipolarity]

Yeah, so basically, the closest the sequences ever gets to 0 is in $|a_0|$. After that it gets further and further away.

So, what is the definition of $\lim_{n\rightarrow +\infty}{a_n}=0$?? Can you find an ε that works?

For all positive ε, we can find some N such that whenever n>N, then $|a_{n}|<ε$. I am working on an appropriate value for ε as we speak.

BiP

 

[Bipolarity]

Hey micromass, I have an idea.
What if we let $ε = |a_{n-1}|$ ?

Then that means there exists N>0 such that whenever n>N, then it must be the case that
$|a_{n}| < |a_{n-1}|$

But am I allowed to shift the index of the term by 1 so I can show that it contradicts the premise?

BiP

 

[micromass]

For all positive ε, we can find some N such that whenever n>N, then $|a_{n}|<ε$. I am working on an appropriate value for ε as we speak.

BiP

Yeah, that is the definition for $\lim_{n\rightarrow +\infty} a_n=0$. But you want to prove that this is NOT true. So you want the negation.

Hey micromass, I have an idea.
What if we let $ε = a_{n-1}$ ?

BiP

Yeah ok. How would that work?

 

[Bipolarity]

Hey micro I just edited my last post. I think it is right, but I confused on why we are allowed to shift the index by one.

In other words, so far we have shown that if $$\lim_{n→∞} a_{n} = 0$$ then it must be the case that for some N>0 whenever n>N, then $$|a_{n}|<|a_{n-1}|$$ which says that the sequence is decreasing.

But the original constraint of the problem is $$|a_{n}|<|a_{n+1}|$$.

I must have made mistake somewhere or not chosen by epsilon correctly.

BiP

 

[micromass]

Hey micromass, I have an idea.
What if we let $ε = |a_{n-1}|$ ?

Then that means there exists N>0 such that whenever n>N, then it must be the case that
$|a_{n}| < |a_{n-1}|$

But am I allowed to shift the index of the term by 1 so I can show that it contradicts the premise?

BiP

Yeah sure you can shift the index. You are given that $|a_{k+1}|>|a_k|$for ALL k. So just choose k=n-1.

But something bothers me. You have not defined n. You can't work with $\varepsilon = |a_{n-1}|$ because you didn't declare what n is.
Why don't you pick a specific value of n??

 

[HallsofIvy]

Choose an arbitrary n, as micromass suggests, and let $\epsilon= |a_n|$, for that specific n. What happens to the definition of $\lim_{n\to\infty} a_n= 0$ using that $\epsilon$?

 

[bda23]

Why don't we just pick [a][/1] as ε? Then |0-[a][/n]| < [a][/1] for all n > N, a contradiction. Or have I missed something?

 

[bda23]

Ok, somehow the font got messed up, but just to make it clear, [a][/1] is meant to be a1.