Playing with fractions

Let \(n\) be an integer such that \(n \ge 2\).Then if we evaluate \[S = 1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + .......+ \frac{1}{n} \]

Can the expression SS ever be an integer???

#NumberTheory #Fractions #CosinesGroup #Goldbach'sConjurersGroup #TorqueGroup

Note by Eddie The Head
7 years, 2 months ago

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Comments

No, it cannot. Let a be the product of all of odd integers less than or equal to n and let 2k1n<2k2^{k-1}\leq n < 2^k. Then a.2k2.Sa.2^{k-2}.S is not going to be an integer, because every term will be an integer, except a.2k2.12k1a.2^{k-2}.\frac{1}{2^{k-1}}.Thus S cannot be an integer.

Bogdan Simeonov - 7 years, 2 months ago

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Can you please clarify have you taken a=a = ??? You have initially defined aa to be the product of all the integers n\le n.This implies a=n!a = n! You also said that all the terms are integers except a2k22k1\frac{a*2^{k-2}}{2^{k-1}}, but this term equals a/2a/2 = n!/2n!/2. If n>1n>1, n!/2n!/2 is clearly an integer.....

Eddie The Head - 7 years, 2 months ago

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Oh, sorry!I meant product of all ODD integers \leq n.Fixed!

Bogdan Simeonov - 7 years, 2 months ago

No, it can't be. A simpler way to see this would be to write (if possible) the above equation (transposing the leading 1 to the right hand side and then simplifying the sum of the fraction) 3.4..n+2.4.n+2.3.4(n1)=bn!3.4.\ldots .n + 2.4.\ldots n + 2.3.4 \ldots (n-1)=b n!, where bb is an integer. Let pp be the largest prime among 1,2,,n1,2,\ldots, n. Convince yourself that pp does not divide any one of the integers p+1,p+2,,np+1,p+2, \ldots, n. Then the terms in the above equation can be grouped as those which contain pp as a factor and the one which does not. The right hand side (n!n!) definitely contains pp as a factor. This gives a contradiction.

Abhishek Sinha - 7 years, 2 months ago

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Really nice...i think you're logically flawless....my solution involved considering the highest power of 2 and then checking the parities of the numerator and the denominator.....

Eddie The Head - 7 years, 2 months ago
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