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

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 08022017 19:32

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 08022017 20:00
(Original post by TverorSecret)
Hi I attach my work, I have made the assumption that this convergense adheres to Archimeadean's property.
However, noting that , I would much rather go:
Given , take . Then,

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 08022017 21:17
(Original post by crashMATHS)
What do you want to know? I think the proof looks okay.
However, noting that , I would much rather go:
Given , take . Then,
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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 08022017 21:39
(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 :_) 
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 08022017 21:46
(Original post by TverorSecret)
Hi I attach my work, I have made the assumption that this convergense adheres to Archimeadean's property. 
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 08022017 22:51
(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.
(Original post by Zacken)
Yes, that's all fine. 
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 09022017 01:12
(Original post by TverorSecret)
..
You write that but the inequality isn't actually strict; the most you can say is that
(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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 15022017 20:06
Why do you write 1/n in that way? idk just seems strange, although rest seems okay

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 15022017 20:27
(Original post by DFranklin)
Actually, there is one (tiny) issue:
You write that but the inequality isn't actually strict; the most you can say is that
(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.
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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 15022017 21:25
(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]
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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 15022017 21:54
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 nonstrict 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. 
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 15022017 22:08
(Original post by DFranklin)
You'd have to ask him/her. It's not really ideal, but it's not indefensible (if you use nonstrict 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.
<= (5+n) / (2/n^2) then would that be more ideal than how OP posed it? 
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 16022017 00:55
(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?
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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