Let \(n\) be an integer such that \(n \ge 2\).Then if we evaluate
\[S = 1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + .......+ \frac{1}{n} \]
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No, it cannot. Let a be the product of all of odd integers less than or equal to n and let 2k−1≤n<2k. Then a.2k−2.S is not going to be an integer, because every term will be an integer, except a.2k−2.2k−11.Thus S cannot be an integer.
Can you please clarify have you taken a= ???
You have initially defined a to be the product of all the integers ≤n.This implies a=n!
You also said that all the terms are integers except 2k−1a∗2k−2, but this term equals a/2 = n!/2.
If n>1, n!/2 is clearly an integer.....
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…(n−1)=bn!, where b is an integer. Let p be the largest prime among 1,2,…,n. Convince yourself that p does not divide any one of the integers p+1,p+2,…,n. Then the terms in the above equation can be grouped as those which contain p as a factor and the one which does not. The right hand side (n!) definitely contains p as a factor. This gives a contradiction.
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.....
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
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\sum_{i=1}^3
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No, it cannot. Let a be the product of all of odd integers less than or equal to n and let 2k−1≤n<2k. Then a.2k−2.S is not going to be an integer, because every term will be an integer, except a.2k−2.2k−11.Thus S cannot be an integer.
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Can you please clarify have you taken a= ??? You have initially defined a to be the product of all the integers ≤n.This implies a=n! You also said that all the terms are integers except 2k−1a∗2k−2, but this term equals a/2 = n!/2. If n>1, n!/2 is clearly an integer.....
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Oh, sorry!I meant product of all ODD integers ≤ n.Fixed!
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…(n−1)=bn!, where b is an integer. Let p be the largest prime among 1,2,…,n. Convince yourself that p does not divide any one of the integers p+1,p+2,…,n. Then the terms in the above equation can be grouped as those which contain p as a factor and the one which does not. The right hand side (n!) definitely contains p as a factor. This gives a contradiction.
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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.....