# 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 $$S$$ ever be an integer???

4 years, 2 months ago

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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 $$2^{k-1}\leq n < 2^k$$. 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.

- 4 years, 2 months ago

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.....

- 4 years, 2 months ago

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

- 4 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) $$3.4.\ldots .n + 2.4.\ldots n + 2.3.4 \ldots (n-1)=b n!$$, 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.

- 4 years, 2 months ago

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.....

- 4 years, 2 months ago