# Sequence convergence question

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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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#2

(Original post by

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

**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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(Original post by

What do you want to know? I think the proof looks okay.

However, noting that , I would much rather go:

Given , take . Then,

**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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#4

(Original post by

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 :_)

**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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#5

**TverorSecret**)

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

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(Original post by

It means the quantity to the left of := is defined by the quantity to the right - equality by definition, if you like.

**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

Yes, that's all fine.

**Zacken**)Yes, that's all fine.

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#7

(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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#8

Why do you write 1/n in that way? idk just seems strange, although rest seems okay

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#9

(Original post by

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.

**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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#10

(Original post by

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]

**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

**cannot**say that LHS is strictly smaller than the RHS.

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#11

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

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

**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.
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#12

**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.

<= (5+n) / (2/n^2) then would that be more ideal than how OP posed it?

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#13

(Original post by

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?

**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?

**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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