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Student 999
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#1
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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Last edited by Student 999; 1 month ago
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mqb2766
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DFranklin
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#3
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#3
(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
Views: 11
Size: 85.4 KB]()
![Name: IMG_F033C8199DBD-1.jpeg
Views: 10
Size: 117.2 KB]()
Not sure how valid my answers to part 3 and 4 are, also for part 2 deduce would a simple explanation be enough?
e.g. for (i)
is a geometric series with common ratio 
. Since |r| < 1, the series converges with value 
. I've gone for "verbose" there, but you could reasonably shorten it to
is a GP with . |r| < 1, so converges to .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
then (a) S_n is an integer,
(b) 0 < n!e - S_n = a_n < 1
and so
.Also, you instead say: you say: "as
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!
Last edited by DFranklin; 1 month ago
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Student 999
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#4
(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) is a geometric series with common ratio . Since |r| < 1, the series converges with value .
I've gone for "verbose" there, but you could reasonably shorten it to
is a GP with . |r| < 1, so converges to .
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
then
(a) S_n is an integer,
(b) 0 < n!e - S_n = a_n < 1
and so.
Also, you instead say: you say: "as 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!
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)
is a geometric series with common ratio . Since |r| < 1, the series converges with value . I've gone for "verbose" there, but you could reasonably shorten it to
is a GP with . |r| < 1, so converges to .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
then (a) S_n is an integer,
(b) 0 < n!e - S_n = a_n < 1
and so
.Also, you instead say: you say: "as
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!
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?
Last edited by Student 999; 1 month ago
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Student 999
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#5
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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Last edited by Student 999; 1 month ago
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DFranklin
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#6
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#6
(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
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?
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?
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DFranklin
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#7
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#7
(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
Views: 9
Size: 33.4 KB]()
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?
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Student 999
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#8
(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.
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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