Conference rusure::math

        The idea is simple, but takes a few lines to express. Suppose if 
possible that S = sum(m=<i=<n)1/i is an integer. This is equivalent to:

        sum(m=<i=<n) n!/((m-1)!i) == 0 mod n!/(m-1)!

which implies:

(*)        sum(m=<i=<n) n!/((m-1)!i) == 0 mod 2^A

where arity(2,n!/(m-1)!) = A. (That is to say that A is the largest natural 
number such that 2^A divides n!/(m-1)!)

        If A = 0, then n=m & the only solution is with n=m=S=1. Otherwise, A>=1.
Let a = max(arity(2,l) | l drawn from {m,...,n}). Note that A >= a >= 1. Define:

        f(i) = n!/((m-1)!i)/(2^(A-a))

        Now, f(i) is an integer. Further, f(i) is odd <=> arity(2,i) = a. So, 
by the definition of a, there is at least one j in {m,...,n} such that f(j) is 
odd. But (*) is just:

        sum(m=<i=<n) f(i) == 0 mod 2^a

        So by parity, there is another number j' (<> j) in {m,...,n} such that 
f(j') is odd. Now, wlog we take j' is "as close as possible" to j. Formally, if 
j = 2^a(2z+1), say, then wlog we can take j' such that  j' = 2^a(2z'+1), where 
|z-z'| = 1. 

        Consider k = (j+j')/2. k certainly lies in {m,...,n}. The trick is that 
arity(2,k) > a, contradicting the definition of a. So no solution, apart from
the trivial n=m=1.

Andy.

what's arity?  looks like parity without the p ...

/Eric

o.k. so 1/a + 1/(a+1) ... + 1/(a+n) is never an integer.

How about leaving out one of the terms ?  When is that an integer ?  If
never, then we have a stronger statement.  If sometimes, then when ?

Obviously, we can't get an integer by leaving out either of the end terms,
since the remaining sum satisfies the original theorem.

Leaving out a middle term gives us:

1/a + 1/(a+1) ... + 1/k + 1/(k+2) ... + 1/n

We left out 1/(k+1).  By the original theorem, neither the stuff to the left
(ending with 1/k) nor the stuff to the right (starting with 1/(k+2)) is an
integer.  But maybe they add to an integer.


/Eric

I thought there might be a solution that used a smaller hammer than the one I
did.  My proof has 2 cases:

(1) 2m>n: the sum is between 0 and 1, hence not an integer, unless m=n=1.

(2) 2m<=n: n>1, so there is a prime p such that n/2<p<=n; then 1/p appears in
the sum but 1/kp doesn't for k>1.  The sum without the 1/p term, in lowest
terms, doesn't have p in the denominator, so cannot equal (rp-1)/p for any r. 
Therefore adding 1/p won't make it an integer.

    & what of 1315.4?
    
    Andy

Please see my note 1547 for previous dabbling in this area.

/Eric

re .5    
    
>(1) 2m>n: the sum is between 0 and 1, hence not an integer, unless m=n=1.
    
    Define S(n) = sigma(i=n+1,2n) 1/i
    
    It is possible to show that S(n) is strictly monotonically increasing
    and, by comparision with sigma 1/i�, that it is bounded above and hence
    that limit(n->oo) S(n) exists.
    
    Questions:
    
    1. Is there an easier demonstration that 0 < S(n) < 1 ?
    
    2. What is a "closed" form expression for the above limit, if any?
    
    Personally I find this fascinating (Yeah, I guess small things amuse
    small minds...) i.e. that even though H(n) diverges, the second half of
    the series only ever contributes at most 1 (in fact, at most some
    number that is the answer to Q2 and can be readily seen to be a little less
    than 1).
    
    Is this a property of more general series?

>   Define S(n) = sigma(i=n+1,2n) 1/i
    
>   It is possible to show that S(n) is strictly monotonically increasing
>   and, by comparision with sigma 1/i�, that it is bounded above and hence
>   that limit(n->oo) S(n) exists.
    
    I don't quite see the comparison with sigma 1/i�, but I'm sure it's there.
        
>   2. What is a "closed" form expression for the above limit, if any?
    
    	ln(2)
    
>   Personally I find this fascinating (Yeah, I guess small things amuse
>   small minds...) i.e. that even though H(n) diverges, the second half of
>   the series only ever contributes at most [ln(2)]
    
    It's an interesting way of putting it. But is this observation
    fundamentally any different from the standard divergence proof
    where huge collections are necessary to form a partial sum of � ?
    
      John

re .9
    
>>   2. What is a "closed" form expression for the above limit, if any?
>    
>    	ln(2)
    
    Is this obvious? Can you prove it?
    
>    It's an interesting way of putting it. But is this observation
>    fundamentally any different from the standard divergence proof
>    where huge collections are necessary to form a partial sum of � ?
    
    No, I guess not.

        re .10
        
>>>	re .9
>>>    
>>>	>>   2. What is a "closed" form expression for the above limit, if any?
>>>	>    
>>>	>    	ln(2)
>>>    
>>>	    Is this obvious? Can you prove it?
        
        It is H(2n) - H(n) and for large k H(k) is approximately ln k + c
        where c is Euler's constant.  Then just use
        
        	(ln(2n) + c) - (ln(n) + c) = ln 2
        
        Dan

    re .11
    
    That makes it obvious provided that H(n) -> ln n + c is obvious. Is it?
    A more first principles demonstration is embedded below.
    
    re .9
        
>    I don't quite see the comparison with sigma 1/i�, but I'm sure it's there.
    
    S(n) = sigma(i=n+1,2n) 1/i
    
    So S(n+1) - S(n)
    
    = sigma(i=n+2,2n+2) 1/i - sigma(i=n+1,2n) 1/i
    = 1/(2n+2) + 1/(2n+1) - 1/(n+1)
    
      2n+1 + 2n+2 - 2(2n+1)
    = ---------------------
      2(n+1)(2n+1)
    
           1
    = ------------
      2(n+1)(2n+1)
    
    This establishes that S(n) is strictly monotonically increasing (which in
    turn given a limit less than one would establish that S(n) < 1 for all n
    as required).
    
    Hence S(n) = sigma(i=1,n) 1/(2i(2i-1))    checking initial condition
    
    [DIGRESSION:
    
    To establish quickly that the limit is less than 1 i.e. without
    actually finding it, showing where the comparison with sigma 1/i�
    comes in. It is immediately obvious but one just needs to be careful
    with range of summation.
    
    S(n)
    = 1/2 sigma(i=1,n) 1/(i(2i-1))
    = 1/2 ( 1 + sigma(i=2,n) 1/(i(2i-1)) )
    < 1/2 ( 1 + sigma(i=2,n) 1/(i(2i-2)) )
    < 1/2 ( 1 + 1/2 sigma(i=2,n) 1/((i-1)(i-1)) )
    < 1/2 ( 1 + 1/2 sigma(i=1,n) 1/(i*i) )
    < 1/2 ( 1 + 1/2 pi*pi / 6 )
    
    which is around 0.91
    END DIGRESSION]
    
    However there is a more direct way.
    
    S(n)
    = sigma(i=1,n) 1/(2i-1) - 1/(2i)
    = sigma(i=1,2n) (-)^(i+1) 1/i
    
    i.e. the alternating harmonic sum.
    
    It is relatively straightforward then to show that S(n) -> ln 2.
                            

If f(x)=1/x, then H(2n)-H(n) = sum(i in n+1..2n)1/i
= (1/n)sum(i in n+1..2n)f(i/n) -> integral(1..2)f(x)dx.

But the latter is ln(2)-ln(1) = ln 2.

>    1. Is there an easier demonstration that 0 < S(n) < 1 ?

Yes, if n>0: clearly S(n) > 0, and S(n)<= n*1/(n+1) since every term is <=
1/(n+1).