# Homework Help: Quick sequence proof help

1. Oct 29, 2013

### phospho

show if the sequence $x_n$ is bounded and $y_n \rightarrow + \infty$ then $x_n + y_n \rightarrow + \infty$

my attempt

if $x_n$ is bounded then $P \leq x_n \leq Q$ for some $P,Q \in \mathbb{R}$ if $y_n \rightarrow + \infty$ then $\forall M>0$ $\exists N \in \mathbb{R}$ s.t. $\forall n > N \Rightarrow y_n > M$

now we need to prove that $\forall M' > 0$ $\exists N_1 \in \mathbb{R}R$ s.t. $\forall n >N_1 \Rightarrow z_n > M'$ where $z_n = x_n + y_n$

$y_n > M$ therefore $x_n + y_n > M + P$ hence take $M' = M + P$ and we get $z_n = x_n + y_n > M'$ hence $\forall n > N_1$ $z_n > M'$

is this all or would I need to add something

2. Oct 29, 2013

### phospho

I'm also worried about if P is negative, could I just state that $M > P$ ?

3. Oct 29, 2013

### LCKurtz

And, for that matter, both $P$ and $Q$ might be negative. To get around having to deal with those cases, just use that $\{x_n\}$ is bounded means there is $B\in \mathbb R$ such that $|x_n|<B$ for all $n$. That tells you $-B<x_n<B$. Can you get a lower bound for $x_n+y_n$ from that? Then figure out if you can make that greater than M'.

4. Oct 29, 2013

### phospho

the lower bound would be $x_n + y_n > M - B$ no? Also, could I not just state that we choose $M > -B$ seeing as M can be any large number?

Also, I don't quite understand how you got $-B < x_n < B$ if it's bounded surely the upper and lower bounds are not necessarily the same?

5. Oct 29, 2013

### LCKurtz

There is no M in my reply or question.

They don't have to be the same to make the claim there is a $B$ that $|x_n|<B$.

6. Oct 30, 2013

### phospho

Is xn is between -B and B then surely B=B? The lower bound would be xn + yn > yn - B

7. Oct 30, 2013

### LCKurtz

Not sure what you are getting at there, but yes, B=B.

Yes. Now you are trying to choose $n$ large enough so that given any $M$, $x_n+y_n > M$. And you know you can make $y_n$ large. How large does $y_n$ have to be?

8. Oct 30, 2013

### phospho

yn has to be larger than B?

9. Oct 30, 2013

### LCKurtz

Stop guessing and start thinking. You have $x_n+y_n>y_n-B$ and you want $x_n+y_n$ to be greater than $M$. How big does $y_n$ need to be to make that happen?

10. Oct 30, 2013

### converting1

11. Oct 30, 2013

### phospho

$y_n > M + B$?

12. Oct 30, 2013

### LCKurtz

Show me why you say that and what it does for you. Write down the inequalities you have. You are almost done.

13. Oct 30, 2013

### phospho

we have $|x_n| \leq B$ and $x_n + y_n \geq y_n - B$
$\forall M > 0 \exists N \in \mathbb{R}$ s.t. $n > N \Rightarrow y_n > M + B$ so $x_n + y_n > M + B - B=M$

this tell us for $M > 0$ $\exists N \in \mathbb{R}$ s.t.$n>N \Rightarrow z_n > M$ where $z_n = x_n + y_n$ (i.e. z_n -> +infinity)

Last edited: Oct 30, 2013
14. Oct 30, 2013

### LCKurtz

OK, that looks pretty good. Well done.

15. Oct 30, 2013

### phospho

Thank you for help

My lecturer keeps telling us that presentation and our grammar is very important when constructing proofs so could you please read through the entire proof below? (this is for an introductory to analysis class):

As $x_n$ is bounded $\exists B \in \mathbb{R}$ s.t. $|x_n| \leq B$
Suppose $\forall M_1 > 0$ since $y_n \rightarrow \infty$ as $n \rightarrow \infty$ $\exists N \in \mathbb{R}$ s.t. $n > N\Rightarrow y_n > M_1$ where $M_1 = M_2 + B$

as $|x_n| \leq B$ $-B \leq x_n \leq B$ so for $n > N$ $x_n + y_n \geq B + y_n > M_1 + B > M_2 + B - B = M_2$ hence $\forall M_2 > 0$ $\exists N \in \mathbb{R}$ s.t. $n>N \Rightarrow x_n + y_n > M_2$ which is the definition for divergence to $+\infty$ therefore $x_n + y_n \rightarrow +\infty$

Last edited: Oct 30, 2013
16. Oct 30, 2013

### LCKurtz

Yes, I agree with your teacher. The order you write it up isn't necessarily the order in which you analyze the problem.

Here you should put in the steps leading to $-B+y_n \le x_n+y_n$

Instead of writing that line like that say it more like this: Suppose $M>0$. Since $y_n\rightarrow \infty$ there exists $N \in \mathbb{R}$ such that for all $n>N$, $y_n>M+B$.

Now instead of that last two lines you can just say since $n > N$, $x_n+y_n \ge ...$ and put your string of inequalities ending with $>M$.

17. Oct 30, 2013

### phospho

I have edited my previous post if you could take a look at it. Thanks.

18. Oct 30, 2013

### LCKurtz

No, I'm sorry, I don't feel like starting over. You shouldn't make major edits after the post has been replied to. It makes it very hard to follow the thread. Now you have $M_1$ and $M_2$'s floating around and silly things like $|x_n|\le B-B$. It's OK if you don't want to follow my suggestions, but at this point I will let your teacher analyze it. You were so close...

19. Oct 30, 2013

### phospho

OK I'll be sure not to make edits but instead post again, but I figured it's still easy to follow as you quoted me so readers can see what you were asking me to change specifically, and I have changed it. Also, I do not have $|x_n| \leq B-B$ it looks like that because I don't know how to use tex and put "$|x_n| \leq B$ $-B \leq x_n \leq B$" if you quote me you will see the code and that it's separated.

I also think it's really demotivating to a student like myself if you say stuff like "you were so close", it just implies I can almost do it, but I'm not there and that you (assuming you are a teacher), have given up on me because I edited my post instead of replying.

Anyhow, I appreciate the help and I'm happy with my solution and I hope the PHD student who will be marking my work will be also.

Thanks again.

20. Oct 30, 2013

### LCKurtz

Well that helps some.

I'm sorry you feel demotivated. I have spent a lot of time in this thread. When you were finally ready to write it up I gave you clear suggestions how to do it properly. All you needed (and still do need) to do was follow the suggestions. There was something to be learned there.

I frequently don't enter this kind of thread because of the difficulty of getting ideas across online. What could be done in a few minutes in person can take days online, especially when we require the student to do most of the work, giving only general hints and guidelines. Anyway, my real life requires me to leave for the day. I'm sure your teacher will have some suggestions.

21. Oct 30, 2013

### phospho

I especially agree with the bolded part, it's very hard for me to smooth something over online, but sometimes I have no other options but to try.

I really appreciate your help as you've helped me with other questions also. I have reached a proof with which I am happy with (after taking your suggestions). Thanks again.