Trying to solve the final part of a Real Analysis problem where the final part is just finding the limit of a sequence. However, the sequence is not explicitly given; relations given to different terms of the sequence are given/shown to be true in previous steps, but you don't have an explicit formula. How do I do this?

Question

mendicant_bias · Answer

Information known:
$$x_{n+1}=\frac{1+x_n}{2}, \ \ \ x_1 > 1.$$$$(\forall n \in N), \ \ \ x_n > 1$$$$(\forall n \in N), \ \ \ x_{n+1}<x_n.$$

So from the above, it's clear that the sequence is monotone decreasing. But how do I find its limit if I'm not given an explicit formula for the sequence? @ganeshie8

ganeshie8 · Answer

What does monotone convergence theorem say ?

mendicant_bias · Answer

Somewhere in my notes I have a quantifier def. version of it, I'll try to find it real quick

ganeshie8 · Answer

No wait, I'll just state it : 

If a sequence of real numbers is decreasing and bounded below, then it has a finite limit, which equals the infimum.

ganeshie8 · Answer

Since our sequence is decreasing and bounded below by $1$, 
from monotone convergence theorem, we know that the limit exists.

ganeshie8 · Answer

Lets go ahead and find it

ganeshie8 · Answer

Since the limit exists, do you agree that we have : 

$\lim\limits_{n\to\infty}x_{n+1} = \lim\limits_{n\to\infty}x_n$

?

mendicant_bias · Answer

That seems reasonable, yeah

ganeshie8 · Answer

Good. Let  $\lim\limits_{n\to\infty}x_{n+1} = \lim\limits_{n\to\infty}x_n=L$

$x_{n+1}=\frac{1+x_n}{2}$

as $n\to\infty$, we get : 
$L=  \frac{1+L}{2}$

$2L = 1+L$

$L=1$

mendicant_bias · Answer

Ah, cool! Hah, I've never thought about it like that, that's awesome!

mendicant_bias · Answer

Thank you.

ganeshie8 · Answer

np : )

Did you figure out the induction proof the other day ?

mendicant_bias · Answer

I was just looking at that, and it made sense to me, but let me double check and just think about it for a minute while I have the chance.

mendicant_bias · Answer

Yeah, part i (x_1 > 1) is given, assume x_k > 1, x_k+1 = 1+x_k all over 2, and since you know x_k is greater than one, (1+x_k)/2 > 1+1/2=1, x_k+1 > 1.

ganeshie8 · Answer

Looks perfect!

mendicant_bias · Answer

Thanks, we'll see how this test coming up goes...I frankly just didn't put in the necessary time prior for all the subjects we're going to cover and I'm just going to have to study what I can in the remaining time.

ganeshie8 · Answer

Looks you're doing great with analysis!  but I think real analysis requires more attention compared to other subjects...  its not that easy even though the topics might seem all familiar...

mendicant_bias · Answer

Yeah, I always thought all my previous math classes were hard and then I ran into analysis which pretty much consists entirely of proofs and there's no neat formulas to find things out, just really being clever and guessing right, it's different from any other math course I've taken so far in how problems are solved.

ganeshie8 · Answer

It's like learning a new language. so much terminology to keep track of... but once you get some hang of it and if you have the analysis blood in you, you're going to love the subject ;)

ganeshie8 · Answer

these lectures helped me a lot with the subject initially https://www.youtube.com/playlist?list=PL04BA7A9EB907EDAF take a look when free..