#sequences calculus 2
rose ermineBOT
lime coral
Is this the correct approach to this problem?
manic trout
I don't see how that helps you. Take an easier example such as the limit of the sequence (1/1, 2/2, 3/3, ...). In that case, arbitrary term b_n is expressed as n/n. Thus, lim as n approaches infinity of b_n just becomes undefined when you express it as
(lim as n approaches infinity of n)/(lim as n approaches infinity of n).
In other words, lim (n/n) = 1, but if we say lim(n/n) = lim(n)/lim(n), then we don't get anywhere.
manic trout
I made up a new sequence to show how you were splitting a convergent sequence into two divergent sequences.
What do you think n/((n+1)^n) approaches?
manic trout
Yeah, so what would y be, and can you take ln of y? Yeah. As for proving it, some rigor can be used, but we kind of know the answer.
manic trout
y is not 1
lime coral
I just wrote y in there
By taking ln of both sides
I’m not sure
But I would assume 0
manic trout
You assumed the limit existed, called the limit y, and then you took ln(y), but is ln(y) defined?
manic trout
Try computing some of the further out terms, such as a_100.
lime coral
It’s just a really small number
manic trout
I'm thinking we could use the squeeze theorem to prove what the limit is.
manic trout
We could take that route too.
manic trout
That's the idea, but you need to do it for a general pair of sequential terms.
lime coral
That was a1 a2 and a3 for the original equation
I get the Socratic questioning, but clearly I am not understanding something, throw me a bone here
Forget the ln of both sides
manic trout
Do you know how to use a spreadsheet?
We can do that to get a sense of the sequence. They talk about things going to infinity "faster" than others, but that is made rigorous with epsilon-delta proofs, but I'm not sure at what level of understanding your instructor wants you to demonstrate the answer.
lime coral
Just write whether it converges or diverges and if it converges, to what number
It’s calc 2
manic trout
I think your idea of showing it to be monotonic and bounded could be an efficient way.
You can write out a_n, and a_(n+1), and compare them with the less than or equals sign.
manic trout
You could show that a_(n+1) <= a_n.
lime coral
Ok well that is true
So now what
It’s a decreasing monotonic sequence
So negative infinity?
That doesn’t seem right
manic trout
That's not what the theorem says.
There's also a statement in the theorem about the sequence being bounded below, and if those two conditions hold, we can say something about the limit.
lime coral
The limit is 0
I can’t see how it approaches any other vslue
Proving it is obviously the tricky part writing it out
manic trout
Do they want you to prove it?
lime coral
No but what good would it be if I didn’t prove it, all my notes I prove my work
An approach like this makes sense to me
That’s why I wrote out the problem the way I did in the first place
manic trout
OK, so there is a theorem that says something about the limit of a sequence that's bounded and only gets closer to the bound and never exceeds it. For example, the sequence (1.5, 1.25, 1.125, 1.0625, ...) is a decreasing sequence but it never goes below 1. I'm just using 1 + 1/n as the general term in this sequence I just defined, starting an n=1.
Taking the log worked there, but taking the log doesn't always work.
lime coral
Let me ask you this, is the problem n/(n+1)^n approaching 0 or not
manic trout
It approaches 0, yes.
lime coral
And how Would you go about proving that
manic trout
You could show that the sequence is strictly decreasing by showing that for any term a_n, the next term, a_(n+1) is smaller, and you could also show that every term is greater than 0. There is then a theorem about monotone sequences you can use: https://en.wikipedia.org/wiki/Monotone_convergence_theorem#Lemma_2
lime coral
So the ratio test
Is probably the way to go about it
Now I know a_n+1 < an
a_n+2 < a_n+1
It’s decreasing
manic trout
You only calculated the ratio for two pairs of terms. To use the ratio test, you would need to calculate the ratio for a general, arbitrary pair of sequential terms.
lime coral
Dude it’s been 2 hours, at that point you could of just written out the problem and explained how you got the answer, I’m not trying to get correct answers to homework problems like everyone else here, I’m trying to understand how to go about the problem, the faster the better, as far as I’m concerned, in a problem as basic as this one where you can clearly see it approaches 0, the ratio test for the first 2 pair of terms is more than enough proof
Sure it won’t work like that for all problems
But I’ll cross that road when I get there
This is literally introduction to sequences that I’m doing
manic trout
I see an answer but it's a bit strange so I'm not sure how to write it out rigorously. If we can show n^2 <= (n+1)^n for all natural numbers greater than 1, the rest is easy, but we need the binomial theorem to do that.
Basically, if n is say, 5, then n^2 = 5^2, but (n+1)^n is 5^5 plus some other stuff, and in each case, the same sort of thing happens. It's sort of obvious that n^2 <= n^n for n greater than 2, but rigorously showing that is just a sort of dull exercise.
lime coral
I wouldn’t stress it, I got other work to do, but atleast your initial point helped me to see that it was 0, so thanks for that
manic trout
It's been a while since I've looked at this stuff so I was looking at the rigorous reasons from the real number axioms for why this goes to 0.
lime coral
Alright well thanks anyways, Imma archive this before it turns into something else lol
manic trout
OK, I apologize that I couldn't help you get a completely rigorous answer.