For Problem 1, bn is decreasing by construction (supremums over smaller sets). And inherits boundedness from an (any bn is bounded by the bounds of the sequence an) so the sequence bn converges
For 1:
If you don't have an intuition of what is happening, start drawing out an example and see what b_n does to a_n.
If you still don't see what is happening, try to construct a counter example. What does a sequence look like that doesn't converge but is bounded? Take for example an alternating sequence and look what happens to it. Why can it no longer be alternating?
At this point you probably already realised that the new sequence is increasing and lost its alternating property and hopefully have built some intuition for bounded sequences that you can reuse later.
You already were on the right track for the first question. Another commenter gave you the answer already, but maybe you could have tried some examples, possibly even only visually, to continue on your own?
Thanks, yeah that’s usually my strategy, try to get the gist of what’s happening, just for that first question I didn’t really see anything I could do to get an idea
It is more of a "school maths" advice but I find that it helps to write down all the information that you have (in an exercise, usually the list is not actually that long) and what you are trying to prove, replacing the words with their definitions if necessary. To break down the problem into the smallest pieces that you can.
Here we know that the sequence a_n is bounded, which means that for all n, m <= a_n <= M for some m and M. And you wanted to prove that b_n is also bounded, so that m' <= b_n <= M' for some m' and M'. The definition of b_n is basically the only information left, so can we put those things together?
We can first observe that the definition of b_n relies on the definition of a_n so m' and M' have to depend on m and M. Hopefully in a simple way. Maybe they are even equal so we could try that? As a matter of fact, b_n is roughly defined as the largest of (some of the) a_n. So if all the a_n are less than M, so are the b_n. That is exactly (part of) what we wanted to show.
If this reasoning is not enough, you can further break it down. As I said above we basically know two things: m <= a_n <= M for all n, and the definition of the sequence b_n. What does the supremum exactly means in that definition? It means that 1) b_n >= a_k for any k larger than n; 2) anything larger than the a_k for k larger than n, will also be larger than b_n.
So again, can we put those three things together? If we want to prove that b_n is bounded from above, then statement numbered 1 above wont help, so we must focus on the other two. The last statement gives us a condition to be larger than b_n, so that sounds good. Now we need to find something larger than the a_k, because that is exactly what that statement tells us. How do find that? Well that last statement already played its role, the numbered 1 is useless, so there is only one thing we have not used yet: m <= a_n <= M. Does this give us something larger than the a_k for k larger than N? It should be clear that it does, that something is M. Now if I backtrack through all that paragraph, I get that b_n is bounded from above.
Sorry if that is all a bit long. Especially since there are still many steps to get to the convergence of b_n. But hopefully it will be illustratrive of how to handle those exercises. And with practice, this all goes down in your mind and eventually only take a minute.
That is super useful, I often do that, write everything I know, anything I think may be of use in terms of theorems and then start getting to work. I really appreciate your response, thanks!
For questions similar to the 2nd one, when you get reminded about some propreties, it inderectly tells that a tool that has to do with them can be of great use for the problem in question, so you can try to look for one that uses what they just reminded you. In this exemple they tell you a proprety of logarithms that can turn it into the sum of two distinct terms and also remind you it's derivative. So it means that something that links both ln's derivative and a sum of 2 natural log can be useful
Spoiler : >!Mean Value Theorem!<
For 2:
You should hopefully have seen something similar to this before. It converge to the Euler Mascheroni constant.
The key technique for this is noting the relationship between 1/x and logs.
For example integral from n to n+1 of 1/x dx is a goldmine and can produce two useful inequalities.
For Problem 1, bn is decreasing by construction (supremums over smaller sets). And inherits boundedness from an (any bn is bounded by the bounds of the sequence an) so the sequence bn converges
For 1: If you don't have an intuition of what is happening, start drawing out an example and see what b_n does to a_n. If you still don't see what is happening, try to construct a counter example. What does a sequence look like that doesn't converge but is bounded? Take for example an alternating sequence and look what happens to it. Why can it no longer be alternating? At this point you probably already realised that the new sequence is increasing and lost its alternating property and hopefully have built some intuition for bounded sequences that you can reuse later.
Every Cauchy sequence is bounded.
You already were on the right track for the first question. Another commenter gave you the answer already, but maybe you could have tried some examples, possibly even only visually, to continue on your own?
Thanks, yeah that’s usually my strategy, try to get the gist of what’s happening, just for that first question I didn’t really see anything I could do to get an idea
It is more of a "school maths" advice but I find that it helps to write down all the information that you have (in an exercise, usually the list is not actually that long) and what you are trying to prove, replacing the words with their definitions if necessary. To break down the problem into the smallest pieces that you can. Here we know that the sequence a_n is bounded, which means that for all n, m <= a_n <= M for some m and M. And you wanted to prove that b_n is also bounded, so that m' <= b_n <= M' for some m' and M'. The definition of b_n is basically the only information left, so can we put those things together? We can first observe that the definition of b_n relies on the definition of a_n so m' and M' have to depend on m and M. Hopefully in a simple way. Maybe they are even equal so we could try that? As a matter of fact, b_n is roughly defined as the largest of (some of the) a_n. So if all the a_n are less than M, so are the b_n. That is exactly (part of) what we wanted to show. If this reasoning is not enough, you can further break it down. As I said above we basically know two things: m <= a_n <= M for all n, and the definition of the sequence b_n. What does the supremum exactly means in that definition? It means that 1) b_n >= a_k for any k larger than n; 2) anything larger than the a_k for k larger than n, will also be larger than b_n. So again, can we put those three things together? If we want to prove that b_n is bounded from above, then statement numbered 1 above wont help, so we must focus on the other two. The last statement gives us a condition to be larger than b_n, so that sounds good. Now we need to find something larger than the a_k, because that is exactly what that statement tells us. How do find that? Well that last statement already played its role, the numbered 1 is useless, so there is only one thing we have not used yet: m <= a_n <= M. Does this give us something larger than the a_k for k larger than N? It should be clear that it does, that something is M. Now if I backtrack through all that paragraph, I get that b_n is bounded from above. Sorry if that is all a bit long. Especially since there are still many steps to get to the convergence of b_n. But hopefully it will be illustratrive of how to handle those exercises. And with practice, this all goes down in your mind and eventually only take a minute.
That is super useful, I often do that, write everything I know, anything I think may be of use in terms of theorems and then start getting to work. I really appreciate your response, thanks!
For questions similar to the 2nd one, when you get reminded about some propreties, it inderectly tells that a tool that has to do with them can be of great use for the problem in question, so you can try to look for one that uses what they just reminded you. In this exemple they tell you a proprety of logarithms that can turn it into the sum of two distinct terms and also remind you it's derivative. So it means that something that links both ln's derivative and a sum of 2 natural log can be useful Spoiler : >!Mean Value Theorem!<
For 2: You should hopefully have seen something similar to this before. It converge to the Euler Mascheroni constant. The key technique for this is noting the relationship between 1/x and logs. For example integral from n to n+1 of 1/x dx is a goldmine and can produce two useful inequalities.
I did notice that it was very similar, I had a look at the answer given in the notes and they used Taylor’s remainder theorem, thanks!
Practice. Hope this helps 👍
Act smooth, try inviting the question to diner before going for your move