Conference rusure::math

If we abort the infinite series

	1/2 + 1/3 + 1/4 + 1/5  . . .

anywhere, can the partial sum ever be an integer ?

>	One set of numbers that reveals solutions are the perfect numbers,
>	namely those the sum of whose factors adds to the number.  For example,
>	6=1+2+3 and 28=1+2+4+7+14.  Note, by the way that 6 in binary is 110
>	and 28 is 11100.  One can prove that similarly, 1111000 is perfect,
>	as is 111110000 etc. etc.

6 = 3 * 2 = (2^2 - 1)(2^(2 - 1))	110 in binary		is perfect
28 = 7 * 4 = (2^3 - 1)(2^(3 - 1))	11100 in binary		is perfect

but

120 = 15 * 8 = (2^4 - 1)(2^(4 - 1))	1111000 in binary	is not perfect

It's divisors are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60,
and 120 which sum to 360, not 240, so 120 is not perfect.  However, the
sum of the reciprocals of those numbers is 3.

n = (2^p - 1)(2^(p - 1)) is a perfect number if and only if 2^p - 1
is prime.  Where 2^p - 1 is the prime q, the sum of the divisors of
n is (1 + q)(1 + 2 + 4 + ... + 2^(p - 1)) = (1 + q)(2^p - 1) = 2^p (2^p - 1)
= 2n.  Dividing through by n means the sum of the reciprocals of the
divisors is 2.  The sets you gave sum to 1 because you left out the
term 1/1.

For n = (2^m - 1)(2^(m - 1)) where 2^m - 1 is not prime, the sum of the
divisors is (sum of divisors of 2^m - 1) * (1 + 2 + ... + 2^(m - 1))
= (sum of divisors of 2^m - 1) * (2^m - 1).  For m = 4 and n = 120, you
luck out that (sum of divisors of 2^m - 1) is a multiple of 2^(m - 1),
making the overall sum a multiple of n.  That doesn't always happen.
31 * 16 is perfect, but the sum of the divisors of 63 * 32 is 13/4.

Dan

By the way, there are infinitely many solutions, as

 1     1      n+1       n        1
--- - --- = ------ - ------ = ------
 n    n+1   n(n+1)   n(n+1)   n(n+1)

So 1/n = 1/(n+1) + 1/(n(n+1)), which means you can take any solution
and expand the last term into a sum of two reciprocals.

Dan

>If we abort the infinite series
>	1/2 + 1/3 + 1/4 + 1/5  . . .
>anywhere, can the partial sum ever be an integer ?

No.

We can rewrite the above series, terminated at the kth term, as

	k!/2 + k!/3 + k!/4 + ... + (k-1)!
	---------------------------------
			k!

Now consider the largest prime p such that p <= k. P obviously divides the
denominator of this fraction, and all the terms in the numerator *except
the pth*. So the fraction cannot be reduced to an integer, since p cannot
divide the numerator. [This result depends on the Lemma that between p and 
p^2 there is another prime q>p.]

Since the above (Harmonic) series diverges, and very slowly, the partial 
sums step past every positive integer, arbitrarily closely, without ever 
equalling any of them.

     <<< Note 1547.4 by CIVAGE::LYNN "Lynn Yarbrough @WNP DTN 427-5663" >>>
                             -< Missed again ... >-

>> Since the above (Harmonic) series diverges, and very slowly, the partial 
>> sums step past every positive integer, arbitrarily closely, without ever 
>> equalling any of them.

At least, every *sufficiently large* positive integer.  (e.g., H(n) differs
from 1 by at least ((1/2 + 1/3 + 1/4) - 1) .)

.2> n = (2^p - 1)(2^(p - 1)) is a perfect number if and only if 2^p - 1
.2> is prime.
    
    Note it is still an open problem if there are any odd perfect numbers.
    (Misreading the above sentence might lead one to assume all perfect
    numbers are of the form given).
    
      John

>>> Since the above (Harmonic) series diverges, and very slowly, the partial 
>>> sums step past every positive integer, arbitrarily closely, without ever 
>>> equalling any of them.
>
>At least, every *sufficiently large* positive integer.  (e.g., H(n) differs
>from 1 by at least ((1/2 + 1/3 + 1/4) - 1) .)


I don't know why you want to add "sufficiently large".  Supposedly, the given
proof showed that the series stops past EVERY positive integer, not just
SUFFICIENTLY LARGE ones.  Please explain.

Thanks.

/Eric

> I don't know why you want to add "sufficiently large".  
    
    Meaning you can only get H(n) within epsilon of every positive integer
    past a certain point (which is a function of epsilon).
    
    You can't get H(n) within say 6.02E-23 of \every/ positive integer; but
    you CAN get within 6.02E-23 of every integer beyond some value N.
    
      John

Mr Lynn wrote (in 1547.4) :


	k!/2 + k!/3 + k!/4 + ... + (k-1)!
	---------------------------------
			k!

Now consider the largest prime p such that p <= k. P obviously divides the
denominator of this fraction, and all the terms in the numerator *except
the pth*. So the fraction cannot be reduced to an integer, since p cannot
divide the numerator. [This result depends on the Lemma that between p and 
p^2 there is another prime q>p.]






I don't see why it depends on the lemma that there's a prime between p
and p^2.

/Eric

    re .10
    
    I haven't been following the details of the proof but...
    
    We are given that p is the largest prime p such that p <= k.
    
    Suppose now that 2p <= k. It is now a false statement that p divides
    all the terms in the numerator except the pth. (In fact then p divides
    all the terms in the numerator but this is not sufficient to show that
    the expression is an integer since p is just one divisor of the
    denominator. So the proof would be flawed if 2p <= k.)
    
    On the other hand if k < 2p then the line of reasoning is correct.
    
    Thus IMHO you are correct to highlight an error in the note. It
    apparently should have said...
    
    "[This result depends on the Lemma that between p and 2p there is
    another prime q > p.]"
    
    and I believe that this Lemma can indeed be proved.
    

>Note 1547.0
>HANNAH::OSMAN
>
>For what sets of distinct integers does the sum of their reciprocals
>add to an integer?
    
	This topic is known as "Egyptian Fractions", apparently due to the
    pyramid-builders enthusiasm for this narrow subject.
    
    	For which +ve integers S does there *not* exist a partition of S
    into +ve integers a1...an, such that sum(i)ai = S, and sum(i)1/ai = 1?
    
    	Same question, but now all the ai must be *distinct*.
    
    The answers to these questions are known, and not too hard to puzzle
    out.
    
    Cheers from wet Geneva,
    Andy.

What is +ve ???

    Eric,
    
    	I'm writing on a rather odd keyboard. I've no idea how the
    characters are visible elsewhere. By "+ve" I meant "positive".
    
    Cheers,
    Andy.

On my screen (decwindows notes running on NT excursion) the "+ve" looks like
a plus sign then a lowercase roman numeral 5 then the base of the natural
logarithm.

Is that what you intended ?  If so, why does that mean "positive" ?

/Eric

    re .15
    
    It's just a convention.
    
    +ve means positive
                    ^^
    
    -ve means negative
                    ^^
    
    You are reading them correctly.

    re .12
    
    I guess that if S is a perfect number then a partition of S into its
    divisors will give an integer sum of reciprocals (2 if you include 1 as
    a divisor and 1 if you exclude 1).
    
    The requirement for distinct a[i] would imply that S must not be a
    perfect square.

    re .17
    
>    I guess that if S is a perfect number then a partition of S into its
>    divisors will give an integer sum of reciprocals (2 if you include 1 as
>    a divisor and 1 if you exclude 1)
    
    Numbers of the form (2^k-1)(2^(k-1)) certainly have the required
    property, but the odd factor does not have to be prime, ie the number
    does not have to be perfect.
    
    Andy.