Zlatman
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HSC 2015 MX2 Marathon (archive) (5 Viewers)
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glittergal96
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Re: HSC 2015 4U Marathon
Let m be the largest integer such that 2^m =< n.
We must then have 2^m =< n < 2^{m+1}, which implies that 2^m > n/2.
Hence k.2^m > n for any positive integer k larger than 1.
Another way of saying this is that 2^m is the only integer in {1,2,...,n} that is divisible by 2^m.
So, assuming that H_n=N is an integer, we have:
Any sum of fractions can be written as a single fraction with denominator the LCM of the denominators of the individual fractions, and the largest power of 2 that occurs in the denominators on the LHS is 2^{m-1}. This implies that
with b odd.
Multiplying both sides by
we get an integer on the LHS and a half integer (number of the form k +`1/2) on the RHS.
This contradiction completes the proof.
An alternate solution to iii):
 (advanced) Prove similarly or otherwise, that$ \ \sum_{i=1}^n \frac{1}{i} \ $is never an integer for$ \ n > 1 )
Let m be the largest integer such that 2^m =< n.
We must then have 2^m =< n < 2^{m+1}, which implies that 2^m > n/2.
Hence k.2^m > n for any positive integer k larger than 1.
Another way of saying this is that 2^m is the only integer in {1,2,...,n} that is divisible by 2^m.
So, assuming that H_n=N is an integer, we have:
Any sum of fractions can be written as a single fraction with denominator the LCM of the denominators of the individual fractions, and the largest power of 2 that occurs in the denominators on the LHS is 2^{m-1}. This implies that
with b odd.
Multiplying both sides by
This contradiction completes the proof.
glittergal96
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Re: HSC 2015 4U Marathon
An elementary proof of divergence can be obtained by writing:
1+1/2+1/3+...
=1+(1/2)+(1/3+1/4)+(1/5+1/6+1/7+1/8)+...
>1+1/2+2/4+4/8+...
=1+1/2+1/2+....
from which we see that the partial sums increase without bound.
Hence? These facts are pretty unrelated, I would be surprised if there was a proof of divergence that used in an essential way that the H_n are non-integers.
An elementary proof of divergence can be obtained by writing:
1+1/2+1/3+...
=1+(1/2)+(1/3+1/4)+(1/5+1/6+1/7+1/8)+...
>1+1/2+2/4+4/8+...
=1+1/2+1/2+....
from which we see that the partial sums increase without bound.
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Ekman
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Re: HSC 2015 4U Marathon
and so if 
the right hand side of the question is true.
For your last statement written in latex, x^5 / 5 is not greater than x^5/3 . Its the other way around actually. But it doesn't really matter either way because it works since you can say:
Zlatman
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Re: HSC 2015 4U Marathon
Fixed, thanks!
oh, lol, that was a typo, should've been the other way round.For your last statement written in latex, x^5 / 5 is not greater than x^5/3 . Its the other way around actually. But it doesn't really matter either way because it works since you can say:
Fixed, thanks!
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