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 \(S\) ever be an integer???

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## Comments

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TopNewestNo, it cannot. Let a be the product of all of odd integers less than or equal to n and let \(2^{k-1}\leq n < 2^k\). Then \(a.2^{k-2}.S\) is not going to be an integer, because every term will be an integer, except \(a.2^{k-2}.\frac{1}{2^{k-1}}\).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 \(\le n\).This implies \(a = n!\) You also said that all the terms are integers except \(\frac{a*2^{k-2}}{2^{k-1}}\), 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 \(\leq\) n.Fixed!

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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) \(3.4.\ldots .n + 2.4.\ldots n + 2.3.4 \ldots (n-1)=b n!\), where \(b\) is an integer. Let \(p\) be the largest prime among \(1,2,\ldots, n\). Convince yourself that \(p\) does not divide any one of the integers \(p+1,p+2, \ldots, n\). 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.....

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