Series with e

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Student 999
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#1
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Not sure how valid my answers to part 3 and 4 are, also for part 2 deduce would a simple explanation be enough?Name:  Screenshot 2022-07-01 at 13.45.06.png
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mqb2766
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Struggling to read your writing in places, but it looks about right.
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DFranklin
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(Original post by Student 999)
Not sure how valid my answers to part 3 and 4 are, also for part 2 deduce would a simple explanation be enough?Name:  Screenshot 2022-07-01 at 13.45.06.png
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As a general comment, try to think why you're saying something is true, and to express that in your argument.
e.g. for (i) b_n is a geometric series with common ratio r = \dfrac{1}{n+1}. Since |r| < 1, the series converges with value \dfrac{1}{n+1}\dfrac{1}{1-1/(n+1)} = \dfrac{1}{n+1}\dfrac{n+1}{n} = \dfrac{1}{n}.

I've gone for "verbose" there, but you could reasonably shorten it to
b_n is a GP with r = \dfrac{1}{n+1}. |r| < 1, so b_n converges to \dfrac{1}{n+1}\dfrac{1}{1-1/(n+1)} = \dfrac{1}{n+1}\dfrac{n+1}{n} = \dfrac{1}{n}.

In (iii), to my mind you've gone astray - it's possible what you meant was valid, but it's really hard to take what you've written and interpret it in a way that makes sense.

They key observation here is that if we define

S_n = \dfrac{n!}{2!} - ... \dfrac{n!}{n!} then

(a) S_n is an integer,
(b) 0 < n!e - S_n = a_n < 1

and so S_n = \lfloor n!e \rfloor.

Also, you instead say: you say: "as 0&lt;a_n&lt;1 it's irrational and not an integer" (emphasis mine). I don't see how you've deduced this and to my mind it's an invalid deduction. It is absolutely true that a_n cannot be an integer, but why couldn't it be rational, say a_n = 1/2?

(iv) I don't think this works because you're using the "a_n is irrational" result that I don't think you've proved sufficiently.

I don't think you can reasonably answer (iv) without contradiction. But if you assume e = m / n with m, n integers, deducing a contradiction from results (ii) and (iii) is immediate.

Edit: couple of other thoughts:

Something you need to be careful with in proofs is an argument that (to an observer) looks like "I'd really like XYZ to be true, so I'll say it's true". What you've done in (iii) really looks like this, whether or not that's what was actually going on - and it's why I think an examiner wouldn't give you many marks there.

From what you've written, it seems possible you think a number that isn't an integer has to be irrational. This is very definitely not the case!
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Student 999
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#4
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(Original post by DFranklin)
As a general comment, try to think why you're saying something is true, and to express that in your argument.
e.g. for (i) b_n is a geometric series with common ratio r = \dfrac{1}{n+1}. Since |r| < 1, the series converges with value \dfrac{1}{n+1}\dfrac{1}{1-1/(n+1)} = \dfrac{1}{n+1}\dfrac{n+1}{n} = \dfrac{1}{n}.

I've gone for "verbose" there, but you could reasonably shorten it to
b_n is a GP with r = \dfrac{1}{n+1}. |r| < 1, so b_n converges to \dfrac{1}{n+1}\dfrac{1}{1-1/(n+1)} = \dfrac{1}{n+1}\dfrac{n+1}{n} = \dfrac{1}{n}.

In (iii), to my mind you've gone astray - it's possible what you meant was valid, but it's really hard to take what you've written and interpret it in a way that makes sense.

They key observation here is that if we define

S_n = \dfrac{n!}{2!} - ... \dfrac{n!}{n!} then

(a) S_n is an integer,
(b) 0 < n!e - S_n = a_n < 1

and so S_n = \lfloor n!e \rfloor.

Also, you instead say: you say: "as 0&lt;a_n&lt;1 it's irrational and not an integer" (emphasis mine). I don't see how you've deduced this and to my mind it's an invalid deduction. It is absolutely true that a_n cannot be an integer, but why couldn't it be rational, say a_n = 1/2?

(iv) I don't think this works because you're using the "a_n is irrational" result that I don't think you've proved sufficiently.

I don't think you can reasonably answer (iv) without contradiction. But if you assume e = m / n with m, n integers, deducing a contradiction from results (ii) and (iii) is immediate.

Edit: couple of other thoughts:

Something you need to be careful with in proofs is an argument that (to an observer) looks like "I'd really like XYZ to be true, so I'll say it's true". What you've done in (iii) really looks like this, whether or not that's what was actually going on - and it's why I think an examiner wouldn't give you many marks there.

From what you've written, it seems possible you think a number that isn't an integer has to be irrational. This is very definitely not the case!
Thanks for the detailed response, for part i I've stated that 1/(n+1) is less than 1 hence convergent or should have been more clearer in showing |r|<1

For part iii I think my idea is essentially the same but haven't presented it well at all and does look like I'm stretching to the answer rather proving it, by introducing a new sum to explain it and argue how after subtracting the integer part of n!e it'll result to a number that isn't an integer which is less than 1 meaning it must be a_n since from part 2 I showed that it's bounded by 0 to 1.

Also in your new sum of s_n why didn't you include the n!/1! as your starting term?

You're correct, I've confused myself by assuming that a_n is irrational just because the expression it asked me to showed contained e which is irrational which is what I should be trying to prove rather than assume?
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Student 999
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For the last part I'm stuck on the contradiction, it doesn't say n tends to infinity so for a small positive integer of n how would you know that q is a factor of n!. I think what you're hinting to is that [n!e]=n!e thus n!e-[n!e] = 0 = a_n hence contradiction but for the above reason how would you deduce that n!e is already an integer for the case that n is small?Name:  IMG_AE417563FE33-1.jpeg
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DFranklin
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(Original post by Student 999)
Thanks for the detailed response, for part i I've stated that 1/(n+1) is less than 1 hence convergent or should have been more clearer in showing |r|<1
My point wasn't that you hadn't proved it, more that you hadn't really explained things (and to point out that you can actually explain and prove quite succinctly if you think about what you want to say as opposed to kind of spewing out maths and leaving it to the examiner to connect the dots).

For part iii I think my idea is essentially the same but haven't presented it well at all and does look like I'm stretching to the answer rather proving it, by introducing a new sum to explain it and argue how after subtracting the integer part of n!e it'll result to a number that isn't an integer which is less than 1 meaning it must be a_n since from part 2 I showed that it's bounded by 0 to 1.
I think the key thing missing in what you wrote is that the n! + n!/2 + ... + n!/n! sum is an integer.

Also in your new sum of s_n why didn't you include the n!/1! as your starting term?
Typo on my part; I originally had it as ... = n! + \dfrac{n!}{2!} + ... which is ... n! + \dfrac{n!}{2!} + ... in TeX format, and then I decided to move it from the RHS to the LHS and just missed the n! term because it was so much smaller than the \dfrac{}{} terms.

You're correct, I've confused myself by assuming that a_n is irrational just because the expression it asked me to showed contained e which is irrational which is what I should be trying to prove rather than assume?
Yeah, don't do that. But if you know it's between 0 and 1 that's enough to deduce it's not an *integer*.
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DFranklin
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(Original post by Student 999)
For the last part I'm stuck on the contradiction, it doesn't say n tends to infinity so for a small positive integer of n how would you know that q is a factor of n!. I think what you're hinting to is that [n!e]=n!e thus n!e-[n!e] = 0 = a_n hence contradiction but for the above reason how would you deduce that n!e is already an integer for the case that n is small?Name:  IMG_AE417563FE33-1.jpeg
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You are assuming (to get a contradiction) that you can write e as m / n for some integers m, n. But then n!e = (n-1)!m which is most definitely an integer.
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Student 999
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#8
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(Original post by DFranklin)
You are assuming (to get a contradiction) that you can write e as m / n for some integers m, n. But then n!e = (n-1)!m which is most definitely an integer.
My bad, thanks you're supposed to choose the specific case of where n is the same as the denominator as the fraction e if it exists since the expression from part 3 holds true for all n.
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