Quick sequence proof help

[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

 

[phospho]

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

 

[LCKurtz]

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

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

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'.

 

[phospho]

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'.

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?

 

[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'.

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?

There is no M in my reply or question.

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?

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

 

[phospho]

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

 

[LCKurtz]

Is xn is between -B and B then surely B=B?

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

The lower bound would be xn + yn > yn - 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?

 

[phospho]

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?

yn has to be larger than B?

 

[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?

 

[converting1]

there's an alternate way if you go about it by contradiction

 

[phospho]

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?

$y_n > M + B$?

 

[LCKurtz]

$y_n > M + B$?

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

 

[phospho]

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

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)

 

[LCKurtz]

OK, that looks pretty good. Well done.

 

[phospho]

OK, that looks pretty good. Well done.

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 $x_n + y_n \rightarrow +\infty$

 

[LCKurtz]

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):

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

As $x_n$ is bounded $\exists B \in \mathbb{R}$ s.t. for all n $|x_n| \leq B$

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

As $y_n \rightarrow \infty$ as $n \rightarrow \infty$ then $\forall M > 0$ $\exists N \in \mathbb{R}$ s.t. $n > N \Rightarrow y_n > M + B$

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 $n>N$, $y_n>M+B$.

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

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 $>M$.

 

[phospho]

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 $>M$.

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

 

[LCKurtz]

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

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...

 

[phospho]

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...

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.

 

[LCKurtz]

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.

Well that helps some.

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.

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.

 

[phospho]

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.

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.