Question

Sequence question
u(n) = u1=1, un = 1/4(u(n-1)^2 + 1, n >=2.

Show that the sequence is increasing and show that it is convergent + calculate its limit.

Can you guide me through the process of doing this please? :)

Answer 1

that would be a 1/4 * (u(n-1))^2 + 1

Answer 2

Where are you stuck? You can show that \(u_n\) is increasing with complete induction.

Answer 3

It is true that \(u_2 \geq u_1\) so you can assume that \(u_n \geq \frac{1}{4}u_{n-1}^2 +1\) for all \(n \in \mathbb{N}\) now you need to show that \(n \leadsto n+1\), how would you do that?

Answer 4

I have no clue ;/ I never really learnt any sequences in school but I have to know these for my exam. I know the basic properties of geometric and arithmetic sequences, that's about it...

Answer 5

ok, I wil try to guide you a bit better then. You have to show that \(u_n \geq u_{n-1}\) this means that the next term in your sequence is always bigger (or equal) to the previous term in your sequence, this is our big goal. You are given a formula for \(u_n\) as in your problem.

It is now my idea to show that by complete induction, I have already shown that it is true for \(n=2\) namely \(u_2 \geq u_1\) . Now you want to assume that \[ u_n \geq u_{n-1}\] and you want to show that this leads to the fact that \(u_{n+1} \geq u_n\) . By complete induction you have shown it entirely then. So what you want do is bring the above in a form you already know, square both sides, divide by 4 and add a one and you will get \[\underbrace{\frac{1}{4}u_{n}^2+1}_{=u_{n+1}} \geq \frac{1}{4}u_{n-1}^2+1= u_n  \implies u_{n+1} \geq u_n \]

Answer 6

Thus your sequence is increasing.

Answer 7

If this does not help you @kingdeolt I can try another approach too

Answer 8

I understand how any n is bigger than n=1, because stuff is just added to it. But your equation is still confusing for me.

Answer 9

ok, consider the following approach then, you should however come back later to this question and try to understand the above (complete induction is one big foundation of mathematics)

You have to show \(u_n \geq u_{n-1}\) this is what it means for a sequence to be increasing, again it only means that the next successful term of the sequence is bigger than the previous one. You want to show that this is always the case in an analytic approach. So consider the following
\[\large u_n \geq u_{n-1} \iff u_n-u_{n-1} \geq 0 \]
Right? I only subtracted \(u_{n-1}\) from both sides, now you are given a formula for what \(u_n\) actually is, substitute it!
\[\frac{1}{4}u_{n-1}^2+1-u_{n-1} \geq 0 \]
this looks like a quadratic equation right? multiply by 4! \[u_{n-1}^2-4u_{n-1}+4 \geq 0 \implies (u_{n-1}-2)^2 \geq 0 \]
all I did was solving the quadratic equation and showing that it can be written as a esquire, however a square is always bigger or equal than zero, thus I have shown that \(u_{n} \geq u_{n-1}\) always is true.

Answer 10

this approach might be clearer and a bit more intuitive @kingdeolt since it only uses basic properties.

Answer 11

Yup, I understood this approach and I will learn the induction when time is available. Now that i've shown it is always increasing, do I use the series formula to try to find an answer, and if I get one, it means it is convergent?

Answer 12

No, think about it as follows. You have shown that \(u_n\) is a sequence that is increasing right? Does this immediately means that \(u_n\) converges? Of course not, it could grow forever right? Mathematically there is no reason why a increasing sequence should stop doing its thing. What could happen?

\(u_n\) gets bigger and bigger and bigger but eventually this giant hits the ceiling (metaphorically) and then it can no longer grow! A bit more rigorously and mathematically speaking would be, \(u_n\) is bound from above, it has an upper bound, a limit that it can not surpass.

So you want to find such a limit, you might want to write down a few terms of your sequence and then take an educated guess of what your limit is.

Answer 13

Did you find a 'guess' for an upper bound? It can be rough.

Answer 14

It is 2, but I did not find it, it was in the answers.

Answer 15

I was thinking the series would probably also approach 2, which would be the limit

Answer 16

no?

Answer 17

Not quite, do you actually have \(u_n= \frac{1}{4}u_{n-1}^2 +1\) or is the \(\frac{1}{4}\) for the entire right hand side?

Answer 18

Lets us show that \(u_n \leq 2\) by complete induction, it is clearly true for \(u_1=1 <2\) so we have our connection, now lets make the hypothesis that \(u_n \leq 2 \) for all natural numbers. So what is left to do is to show that \(u_{n+1} \leq 2\) but we have a formula for \(u_n\) so lets make use of it:
\[\large u_{n+1}= \frac{1}{4}u_n^2+1 \leq \frac{1}{4}2^2+1=2 \implies u_{n+1} \leq 2 \]

Answer 19

now you have shown that \(u_n\) is convergent.

Answer 20

By the way I am not quite sure what you mean with series, you mean taking the sum of the sequence (which is defined as the series)? You cannot argument convergence of a sequence through a series and vice versa, they are to be considered as two independent things.

\(u_n = \frac{1}{n}\) converges but \(\sum_{n=1}^{\infty} \frac{1}{n}\) diverges.

Answer 21

I never thought sequences would be a harder topic than geometry, calculus and probability combined

Answer 22

Always depends on the level which you're studying them at. Or how deep you want to go, the important thing to remember about sequences is

Monotonicity (monotone increasing or decreasing) AND being bound (from above or below) \(\implies\) convergence of a sequence.

And to be absolutely strict about it, convergences means \(\forall \epsilon > 0 \exists n \in \mathbb{N} \) such that \(\forall n_0 \geq n, |u_{n_0}-u_n| \leq \epsilon \)

Answer 23

That last line helped :P. Don't have a clue of what the symbols represent or whatever. Yeah, but at my level... This is hard to get through... ;/

Answer 24

You've been really good and patient though, thanks a lot :)

Answer 25

you're welcome