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???

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

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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.

Bogdan Simeonov - 7 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.....

Eddie The Head - 7 years, 2 months ago

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

Abhishek Sinha - 7 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.....

Eddie The Head - 7 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$