Proving convergence of sequence from convergent subsequences

  • #1
lys04
112
4
In the photos are two proof questions requiring proving convergence of sequence from convergent subsequences. Are my proofs for these two questions correct? Note in the first question I have already proved that f_n_k is both monotone and bounded

Thanks a lot in advance!

IMG_0326.jpeg

IMG_0327.jpeg


IMG_0325.jpeg

IMG_0324.jpeg

IMG_0329.jpeg
 
Physics news on Phys.org
  • #2
You lost me on your first proof. Note that ##f_n## is monotone increasing.
 
  • #3
PS if I asked you to state why you think the first proposition holds, what would you say? Without any calculations, why must ##f_n## converge to ##\frac 1 2##?
 
  • #4
The proposed proofs may be correct, but they certainly can be made more clear.
First, can you show that ##f_n \le 1/2 \ \ \forall n\in \mathbb N##?
Then start with an arbitrary ##\epsilon \gt 0## and see how far you can get.
 
  • Like
Likes PeroK
  • #5
I think OP, we can dismiss the sequence diverging to infinity or by alternating, by being monotone. Additional limit points are out too. Convergence to 1/2 is left.
 
  • #6
PeroK said:
You lost me on your first proof. Note that ##f_n## is monotone increasing.
Where did I lose you? I wrote that f_n is monotone increasing in the first proof?
 
  • #7
lys04 said:
Where did I lose you? I wrote that f_n is monotone increasing in the first proof?
Okay, I can see what you are doing now.
 
Last edited:
  • #8
Proving the set of indices not in the subsequence converges is really roundabout. Given ##\epsilon>0## there exists ##n_k## such that ##|1/2-f_{n_k}|<\epsilon##. What can you say about ##f_n## for any ##n>n_k##? Don't think about whether it's part of the subsequence or not.
 
  • #9
Let [itex]\epsilon > 0[/itex]. Then by convergence of the subsequence there exists [itex]K \in \mathbb{N}[/itex] such that if [itex]k > K[/itex] then [tex]\tfrac12 - \epsilon < f_{n_k} \leq \tfrac12.[/tex] But by monotonicity, we have [itex]f_{n_k} \leq f_n \leq f_{n_{k+1}} \leq \frac12[/itex] for every [itex]n_k < n < n_{k+1}[/itex]. It then follows that [tex]\tfrac12 - \epsilon < f_{n_K} \leq f_n \leq \tfrac12[/tex] for every [itex]n > n_K[/itex], showing that [itex]f_n \to \frac12[/itex].
 
  • #10
Office_Shredder said:
Proving the set of indices not in the subsequence converges is really roundabout. Given ϵ>0 there exists nk such that |1/2−fnk|<ϵ. What can you say about fn for any n>nk? Don't think about whether it's part of the subsequence or not.
fn>fnk since nk is a sequence of increasing numbers?
 
  • #11
pasmith said:
Let ϵ>0. Then by convergence of the subsequence there exists K∈N such that if k>K then 1/2−ϵ<fnk≤1/2.
wait why did the epsilon on the right hand side disappear?
 
  • #12
lys04 said:
wait why did the epsilon on the right hand side disappear?
If any ##f_{n_k}## is larger than 1/2 and they are monotonic increasing, can you prove that they must all be significantly greater than 1/2 from that point on? What does that say about the convergence to 1/2?
 
  • #13
PeroK said:
Okay, I can see what you are doing now.
are they correct?
 
  • #14
FactChecker said:
If any ##f_{n_k}## is larger than 1/2 and they are monotonic increasing, can you prove that they must all be significantly greater than 1/2 from that point on? What does that say about the convergence to 1/2?
f_n_k doesn't converge to 1/2...?
 
  • #15
lys04 said:
are they correct?
I don't think there are any mistakes, but the consensus is that you that they could be simpler. In the second proof, you spend a lot of time unnecessarily abstracting the concept of odd and even numbers. And the main N-epsilon details get omitted somewhat.

For example, for the second one I would do:

Let ##\epsilon > 0##. From the convergence of the odd and even subsequences, there exist ##N_1, N_2## such that if ##n## is odd:
$$n > N_1 \implies |f_n - L| < \epsilon$$And if ##n## is even:$$n > N_2 \implies |f_n - L| < \epsilon$$Let ##N = \max\{N_1, N_2\}##, then:
$$n > N \implies |f_n - L| < \epsilon$$Hence ##f_n \to L##.

My proof is quite formulaic and if someone asked for more details, you can simply add more detail or justification at any step. Whereas, in your proofs, the logic is a little jumbled.

PS also if I did make a mistake in my proof, it would be much easier to identify where one step does not follow from another.

The main point is that as your proofs get more difficult, my approach would generalise better. Whereas, your proofs are likely to become labyrithine. That's something to think about.
 
  • Like
Likes lys04, FactChecker and docnet
  • #16
lys04 said:
f_n_k doesn't converge to 1/2...?
This is a good beginning exercise in formal proofs. Suppose there is an ##f_{n_k} \gt 1/2## and they are monotonic increasing. Can you formally prove that they can not converge to 1/2?
 
Back
Top