TverorSecret
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Hi I attach my work, I have made the assumption that this convergense adheres to Archimeadean's property.

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crashMATHS
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(Original post by TverorSecret)
Hi I attach my work, I have made the assumption that this convergense adheres to Archimeadean's property.
What do you want to know? I think the proof looks okay.

However, noting that  \displaystyle \frac{5}{n} - \frac{2}{n^2} < \frac{5}{n} , I would much rather go:

Given  \varepsilon > 0 , take  \displaystyle N : = \Bigl \lceil \frac{5}{\varepsilon}}  \Bigl \rceil . Then,


 \displaystyle n> N \Rightarrow \Bigl\lvert 5 \left(1+\frac{1}{n} \right) - 2 \left(1+\frac{1}{n^2}\right) - 3 \Bigl\rvert < \frac{5}{n} < \varepsilon
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TverorSecret
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(Original post by crashMATHS)
What do you want to know? I think the proof looks okay.

However, noting that  \displaystyle \frac{5}{n} - \frac{2}{n^2} < \frac{5}{n} , I would much rather go:

Given  \varepsilon > 0 , take  \displaystyle N : = \Bigl \lceil \frac{5}{\varepsilon}}  \Bigl \rceil . Then,


 \displaystyle n> N \Rightarrow \Bigl\lvert 5 \left(1+\frac{1}{n} \right) - 2 \left(1+\frac{1}{n^2}\right) - 3 \Bigl\rvert < \frac{5}{n} < \varepsilon
Sorry I wanted to know if people thought it was correct, as I don't have a model solution to these types of questions.

Yes thank you, I like what you wrote - may I ask what the colon before the equals sign represents in the third line.

Thank you for your response :_)
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crashMATHS
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(Original post by TverorSecret)
Sorry I wanted to know if people thought it was correct, as I don't have a model solution to these types of questions.

Yes thank you, I like what you wrote - may I ask what the colon before the equals sign represents in the third line.

Thank you for your response :_)
It means the quantity to the left of := is defined by the quantity to the right - equality by definition, if you like.
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Zacken
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(Original post by TverorSecret)
Hi I attach my work, I have made the assumption that this convergense adheres to Archimeadean's property.
Yes, that's all fine.
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TverorSecret
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(Original post by crashMATHS)
It means the quantity to the left of := is defined by the quantity to the right - equality by definition, if you like.
Great yeah ic - thanks for your help.

(Original post by Zacken)
Yes, that's all fine.
Awesome thnx
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DFranklin
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(Original post by TverorSecret)
..
Actually, there is one (tiny) issue:

You write that | 5 \frac{1}{n} - 2 \frac{1}{n^2} | < \frac{5}{n}-\frac{2}{n^2} but the inequality isn't actually strict; the most you can say is that

| 5 \frac{1}{n} - 2 \frac{1}{n^2} | \leq \frac{5}{n}-\frac{2}{n^2}

(and in fact you will generally have equality).

It's not a big deal, but it's a good idea to get used to being careful about things like this, sometimes it does become important.
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alexgreyx
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Why do you write 1/n in that way? idk just seems strange, although rest seems okay
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alexgreyx
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(Original post by DFranklin)
Actually, there is one (tiny) issue:

You write that | 5 \frac{1}{n} - 2 \frac{1}{n^2} | < \frac{5}{n}-\frac{2}{n^2} but the inequality isn't actually strict; the most you can say is that

| 5 \frac{1}{n} - 2 \frac{1}{n^2} | \leq \frac{5}{n}-\frac{2}{n^2}

(and in fact you will generally have equality).

It's not a big deal, but it's a good idea to get used to being careful about things like this, sometimes it does become important.
Why did OP do:

5 \frac{1}{n} - 2 \frac{1}{n^2} | < \frac{5}{n}-\frac{2}{n^2}[/latex]

Instead of:

5 \frac{1}{n} - 2 \frac{1}{n^2} | < \frac{5}{n}+\frac{2}{n^2}[/latex]

And yes the inequality should be strict, but for values greater than 0, then LHS = RHS for all values as OP had written
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DFranklin
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(Original post by alexgreyx)
Why did OP do:

5 \frac{1}{n} - 2 \frac{1}{n^2} | < \frac{5}{n}-\frac{2}{n^2}[/latex]

Instead of:

5 \frac{1}{n} - 2 \frac{1}{n^2} | < \frac{5}{n}+\frac{2}{n^2}[/latex]
You'd have to ask him/her. It's not really ideal, but it's not indefensible (if you use non-strict inequality signs).

And yes the inequality should be strict, but for values greater than 0, then LHS = RHS for all values as OP had written
No, it's precisely when you may have that LHS = RHS that you cannot say that LHS is strictly smaller than the RHS.
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alexgreyx
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(Original post by TverorSecret)
Great yeah ic - thanks for your help.



Awesome thnx
Why did you do:

5 \frac{1}{n} - 2 \frac{1}{n^2} | < \frac{5}{n}-\frac{2}{n^2}[/latex]

Instead of:

5 \frac{1}{n} - 2 \frac{1}{n^2} | < \frac{5}{n}+\frac{2}{n^2}[/latex]

(Original post by DFranklin)
You'd have to ask him/her. It's not really ideal, but it's not indefensible (if you use non-strict inequality signs).

No, it's precisely when you may have that LHS = RHS that you cannot say that LHS is strictly smaller than the RHS.
That makes more sense yes.
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alexgreyx
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(Original post by DFranklin)
You'd have to ask him/her. It's not really ideal, but it's not indefensible (if you use non-strict inequality signs).

No, it's precisely when you may have that LHS = RHS that you cannot say that LHS is strictly smaller than the RHS.
So if they had used a + sign s.t. the end of their first line read:

<= (5+n) / (2/n^2) then would that be more ideal than how OP posed it?
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DFranklin
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(Original post by alexgreyx)
So if they had used a + sign s.t. the end of their first line read:

<= (5+n) / (2/n^2) then would that be more ideal than how OP posed it?
Assuming you mean (5+n)/(2n^2), not hugely. You generally want to do two things when saying "A < B" in part of an analysis argument. First you should make sure than it's true that "A < B", but secondly, if you're then going to be working with B, you want to try to make sure B is something nice to work with. (5+n)/(2n^2) passes the first test but not the second.

Probably the cleanest approach here is to go:

for n >= 1, 0 < 2/n^2 < 5/n, so 0 < 5/n - 2/n^2 < 5/n, and then given epsilon > 0, we have n > 5/epsilon => 5/n < epsilon.

But to be honest it's hard to definitively mess this question up - even if you omit justification (as I'd guess is what happened here in the original post), the justification is so trivial that you're likely to get the benefit of the doubt unless someone's being incredibly picky.
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