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If I tell you that: a_n := sup x ∈ (0,∞)xe^{-nx} i.e. a_n is the supremum of the set of values xe^{-nx} for each value of x > 0.

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Then, if this set of values is strictly bounded above by 1/n, why does this imply that the supremum of the set is strictly less than 1/n?

I thought, OK, so let's label the set S = x ∈ (0,∞)xe^{-nx}. Then, we have that all the values in S < 1/n. So, we have that any upper bound of S must be >= 1/n. Hence, any supremum must be smaller than any upper bound, so it must be <= 1/n. So, I had sup(S) = 1/n, and so I wrote sup(S) <= 1/n, and then I used the Sandwich theorem. This was my intuition.

Can someone please help? Thank you
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ghostwalker
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(Original post by Takeover Season)
If I tell you that: a_n := sup x ∈ (0,∞)xe^{-nx} i.e. a_n is the supremum of the set of values xe^{-nx} for each value of x > 0.

Name:  Screen Shot 2020-07-19 at 04.14.43.png
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Then, if this set of values is strictly bounded above by 1/n, why does this imply that the supremum of the set is strictly less than 1/n?
Of itself, it doesn't.

You could actually work out the maximum value of xe^{-nx} over the interval (0,\infty) and it comes to \frac{1}{ne}, and hence &lt;\frac{1}{n}

I thought, OK, so let's label the set S = x ∈ (0,∞)xe^{-nx}. Then, we have that all the values in S < 1/n.
Not sure why you called it by another name, S; a_n seemed fine.

So, we have that any upper bound of S must be >= 1/n.
No, this isn't true. You can only claim this if you know that 1/n is the supremum (least upper bound). And in fact it isn't. It's just an upper bound.
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Takeover Season
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(Original post by ghostwalker)
Of itself, it doesn't.

You could actually work out the maximum value of xe^{-nx} over the interval (0,\infty) and it comes to \frac{1}{ne}, and hence &lt;\frac{1}{n}



Not sure why you called it by another name, S; a_n seemed fine.



No, this isn't true. You can only claim this if you know that 1/n is the supremum (least upper bound). And in fact it isn't. It's just an upper bound.
Hi, thank you for your reply. Sorry, I meant S is the set of values of xe^{-nx} over the interval (0,\infty) and a_n is the supremum of this set of values. Now, we have shown that each value in the set S &lt;\frac{1}{n}.

Oh, so now I understand. I can only say that any upper bound of S \geq either the supremum of S or the maximum of S, and as \frac{1}{n} \neq \frac{1}{ne} = max/sup(S), I cannot make that conclusion.
However, it would've been right instead to say that: any upper bound of S \geq \frac{1}{ne} right?

Oh yes, so that means as each value in the set S is bounded above by &lt;\frac{1}{n}, then it must be that \frac{1}{n} is an upper bound of S. As the supremum of a set S is less than or equal to every upper bound of a set, then a_n \leq \frac{1}{n}.

Then, we have 0 \leq a_n \leq \frac{1}{n} and the conclusion follows.
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ghostwalker
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(Original post by Takeover Season)
Hi, thank you for your reply. Sorry, I meant S is the set of values of xe^{-nx} over the interval (0,\infty) and a_n is the supremum of this set of values. Now, we have shown that each value in the set S &lt;\frac{1}{n}.
Sorry, I missed that distinction. My bad.

Oh, so now I understand. I can only say that any upper bound of S \geq either the supremum of S or the maximum of S, and as \frac{1}{n} \neq \frac{1}{ne} = max/sup(S), I cannot make that conclusion.
However, it would've been right instead to say that: any upper bound of S \geq \frac{1}{ne} right?
Yes.

Oh yes, so that means as each value in the set S is bounded above by &lt;\frac{1}{n}, then it must be that \frac{1}{n} is an upper bound of S. As the supremum of a set S is less than or equal to every upper bound of a set, then a_n \leq \frac{1}{n}.

Then, we have 0 \leq a_n \leq \frac{1}{n} and the conclusion follows.
Well we can go a bit further. Knowing that a_n=\frac{1}{en} we can say a_n&lt;\frac{1}{n} with the strict inequality - in this case. And that justifies its usage in the given text.

But without that, or some similiar working, we can only say a_n\leq \frac{1}{n}
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(Original post by ghostwalker)
Sorry, I missed that distinction. My bad.



Yes.



Well we can go a bit further. Knowing that a_n=\frac{1}{en} we can say a_n&lt;\frac{1}{n} with the strict inequality - in this case. And that justifies its usage in the given text.

But without that, or some similiar working, we can only say a_n\leq \frac{1}{n}
No problem.

Oh right, thank you - that makes perfect sense! Yes, the given working wasn't in the text, so I was confused as to how the strict inequality appeared, but I assume they expect me to do that working out.

Thanks a lot!
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