Question

Let\[x_1,x_2,...\]be a succession of real, non-negative numbers such that\[x_{n+1}\le x_n+\frac{1}{n^2}\]for all\[n\ge1\]Show that\[\lim_{n \rightarrow \infty}x_n\]exists.

Answer 1

Time for breakfast first...

Answer 2

*bookmark (I'll be on later)

Answer 3

ha...trivial...I'll let you think about it :)

Answer 4

That's only fair, y'all basically solved the other two for me. I need to learn to think about these kinds of problems.

Answer 5

try writing out the terms TuringTest - you should see how to solve it then...

Answer 6

use the fact that the MAXIMUM \(x_{n+1}=x_n+\frac{1}{n^2}\)

Answer 7

Okay, this kinda thing is really new to me, so I'm gonna be a bit slow getting it, but thanks for the tip!

Answer 8

what would the first term be?

Answer 9

It doesn't say a specific number, it's just a succession of positive real numbers. So you just mean\[x_2\le x_1+1\]eh?

Answer 10

yes - first term is \(x_1\)
2nd term MAX is \(x_1+\frac{1}{1^2}\)
now, what would the 3rd term MAX be?

Answer 11

obviously\[x_2+\frac{1}{4}\]

Answer 12

don't add up the values - leave them as 1/n^2

Answer 13

and write the MAX 3rd term in terms of \(x_1\)

Answer 14

you should spot a pattern emerging...

Answer 15

ah...
x_3 max is\[x_2+\frac{1}{2^2}=x_1+1+\frac{1}{2^2}\]x_n max is\[x_{n-1}+1+\frac{1}{2^2}+\dots+\frac{1}{(n-1)^2}\]

Answer 16

1st term should be \(x_1\)

Answer 17

yeah, just didn't write it.
\[x_1\]x_2 max is\[x_2=x_1+1\]x_3 max is\[x_2+\frac{1}{2^2}=x_1+1+\frac{1}{2^2}\]x_n max is\[x_{n-1}+1+\frac{1}{2^2}+\dots+\frac{1}{(n-1)^2}\]

Answer 18

you should end up with something like:\[\text{MAX }x_{n+1}=x_1+\sum_1^n\frac{1}{n^2}\]

Answer 19

the sum on the right has a well known value

Answer 20

typo again...
\[x_1\]x_2 max is\[x_2=x_1+1\]x_3 max is\[x_2+\frac{1}{2^2}=x_1+1+\frac{1}{2^2}\]x_n max is\[x_1+1+\frac{1}{2^2}+\dots+\frac{1}{(n-1)^2}\]so that leads to your formula clearly

Answer 21

BTW: I meant the 1st in the last line of your reply should be \(x_1\)

Answer 22

ok - so do you recognise the sum?

Answer 23

yeah, caught that^^^
server is so slow...

Answer 24

remember we want to know what happens as n tends to \(\inf\)

Answer 25

actaully we just need to show that a limit exists - you don't need the value of it

Answer 26

Ok, so the x_n are bounded by \( x_1 + π^2/6 \). Now why does that mean they have a limit?

Answer 27

Well because it's finite of course

Answer 28

the sum is well known and has a value of \(\frac{\pi^2}{6}\) as JamesJ stated

Answer 29

That I couldn't recall, though I knew it was finite

Answer 30

so we now know that:\[x_{n+1}\le x_1+\frac{\pi^2}{6}\]

Answer 31

isn't that enough to say that the limit is finite because x_1 and the sum are finite? regardless of n?

Answer 32

and that the sum exists of course as James pointed out.

Answer 33

maybe this is not a /strong/ enough proof. I thought this was enough but there must be more to it. maybe @JamesJ can help us understand?

Answer 34

The thing is you can have a bounded sequence that doesn't have a limit. For instance
a_n = (-1)^n

What you can say is that a bounded sequence has a convergent subsequence. You might want to try and show the sequence converges to the limit of a subsequence.

But you may not want to go down that road.

Answer 35

Oh man, I don't even know what a subsequence is. Is that the sequence of partial sums?

Answer 36

do we need to use L'Hôpital's rule?

Answer 37

there are no sums here, except in the bound calculation we just did.. You want to show that (xn) converges. Not the sum of (xn) or anything else.

Answer 38

do we just need to show that the ratio between successive terms gets smaller, and therefore it converges?

Answer 39

(And L'Hopital has absolutely nothing to do with it.)

Answer 40

sorry - just grasping at straws here :-(

Answer 41

A strategy would be to show that (xn) is ultimately a Cauchy sequence. Then (xn) converges.

Answer 42

These terms--subseqnece, Cauchy--are all from Real Analysis. Having some knowledge of these things and standard theorems will help.

Answer 43

*subsequence

Answer 44

\[x_{n+1}-x_n\le\frac{1}{n^2}\]and we know 1/n^2 converges - is that sufficient?

Answer 45

the Cauchy sequence approach is what asnaseer said isn't it? Showing that the elements get arbitrarily closer to each other as n increases

Answer 46

If you absolute value then yes, that would show (xn) is Cauchy and then you could conclude (xn) converges. But unfortunately, you don't have
\[ | x_{n+1} - x_n | < 1/n^2 \]

Answer 47

:(

Answer 48

JamesJ - does it have to be strictly LESS THAN? LESS THAN OR EQUAL is not enough?

Answer 49

right because it can be large negative...

Answer 50

less than/less than or equal; doesn't matter. What's important is the absolute value.

Anyway, I think the Cauchy argument is the way to go. Suppose the (xn) is not Cauchy. Write that down in \( \epsilon-N \) terms and derive a contradiction.

Answer 51

ok - thanks - time to go learn about Cauchy...

Answer 52

Yikes, I haven't done that sort of thing yet.
Yes, study time!

Answer 53

which is good - we are increasing our knowledge! :-)

Answer 54

That's why I'm here :-)

Answer 55

ok, I've learned a little about Cauchy Sequences and Real Analysis (quite interesting) and have formulated this so far:\(\{x_n\}\) is Cauchy if for any \(\epsilon>0\) there is a number N such that \(|x_n-x_m|<\epsilon\) for n,m>N.
So we can write:\[\begin{align}
|x_n-x_m|&=|(x_n-x_{n-1})+(x_{n-1}-x_{n-2})+...+(x_{m+1}-x_m)|\\
&\le|x_n-x_{n-1}|+|x_{n-1}-x_{n-2}|+...+|x_{m+1}-x_m|\\
&\le\frac{1}{(n-1)^2}+\frac{1}{(n-2)^2}+...+\frac{1}{m^2}\\
&\le\psi^1(m)-\psi^1(n+1)
\end{align}\]
I'm not sure how to proceed from here.
Am I even on the right track here @JamesJ?

Answer 56

As we commented above, you don't know that
\[ |x_n - x_{n-1} | \leq \frac{1}{(n-1)^2} \]
If you did, this problem would be easy. But alas, we don't.

Answer 57

Why did Zarkon call this problem 'trivial' if it involves so much I wonder...

Answer 58

but we are given:\[x_{n+1}\le x_n+\frac{1}{n^2}\]therefore:\[x_{n+1}-x_n\le \frac{1}{n^2}\]and since the right hand side is always positive aren't we allowed to just add the |'s around the left hand side?

Answer 59

@Zarkon - have we missed one of your BIG FOOTPRINTS somewhere?

Answer 60

there is more than one way to do a problem

Answer 61

It's actually not hard if you know how to write down the converse of the Cauchy condition. But perhaps he has something else in mind.

Answer 62

@Zarkon - I think your reply should have been:
"to do a problem, more than one way, there is"
in the style of Yoda :-)

Answer 63

I'll keep that in mind ;)

Answer 64

can you give us a clue @Zarkon?

Answer 65

@JamesJ - why can't we write the abs value as indicated?

Answer 66

Suppose x_4 = 1. Then
\[ x_5 \leq x_4 + 1/4^2 = 17/16 \]
so \( x_5 \) could be 1/2 say, in which case it is not true that
\[ | x_5 - x_4 | \leq 1/4^2 \]

Answer 67

so for recursive definitions of inequalities - we can't use the same rules as for normal inequalities I guess

Answer 68

consider the sequence defined by

\[y_1=x_1\]

and \[y_{n+1}=x_{n+1}-\sum_{i=1}^{n}\frac{1}{i^2}\]

Answer 69

Nice.

Answer 70

\(y_{n+1}\) is always equal to \(x_1\)?

Answer 71

no

Answer 72

sorry always <= x_1

Answer 73

\[y_{n+1}=x_{n+1}-\sum_{i=1}^{n}\frac{1}{i^2}\le x_{n}+\frac{1}{n^2}-\sum_{i=1}^{n}\frac{1}{i^2} \]

\[=x_{n}-\sum_{i=1}^{n-1}\frac{1}{i^2}=y_n\]

Answer 74

i.e., \( y_{n+1} \leq y_n \). Now ...

Answer 75

so the {\(y_n\)} sequence converges

Answer 76

yes...because it is bonded below and decreasing.

Answer 77

*bounded

Answer 78

and there must be a method to show that the {\(x_n\)} sequence also converges because of this - more reading?

Answer 79

let \[y_n\to y\]

\[\lim_{n\to \infty}x_n=\lim_{n\to \infty}y_{n}+\sum_{i=1}^{n-1}\frac{1}{i^2}=y+\frac{\pi^2}{6}\]

Answer 80

I'll be honest and say I prefer this method to mine because it's direct; vs. mine which would be an indirect proof (and theoretically, a little more difficult).

Answer 81

Well I'm lagging pretty far behind the 3 of you, guess I don't have much to contribute. I did at least follow it though up until that last step. What is y?

Answer 82

thats a nice proof - thanks @Zarkon. and thanks @JamesJ for introducing me to the world of Real Analysis.

Answer 83

the limit of \(y_n\)

we do now know its value

Answer 84

ah, ok, then I guess I get it. I'll have to study it though.
Thanks guys.

Answer 85

The result that was just used is one of the most important basic results in analysis and worth knowing:

Let (x_n) be sequence of real numbers which is
- increasing and bounded above; or
- decreasing and bounded below
then (x_n) has a limit.

Answer 86

thx @JamesJ