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Unable to guess that how the simple reduced form, which I hope is the irreducible form, has an even denominator. It also means that the numerator is odd, while denominator is even.

So, working backwards from the result to be proven,the form is : $\frac{2k'+1}{2k} \exists k,k' \in \mathbb{N}$ and $k' \lt k$ as my argument below shows.

The sum of all terms would have the initial denominator as: $p(p+1)(p+2)...(p+n)$.

Nothing else comes to mind. I hope such type of question needs only induction, and any other approach will not be thinkable even.

Also, here strong form of induction seems the key, although I don't know how to apply that and am trying the simple (weak) form of induction. Hence, the starting case would be the sum of two terms, i.e. $1/p + 1/(p+1) = 2p+1/(p.(p+1))$. The numerator is odd, and the product of two consecutive numbers will be even too. Hence, base case is proved.

Will the assumption for the inductive hypothesis holds for any $n=m$. I feel that there is involved a sort of circular reasoning here, as need to check for all possible values of $m$ that the hypothesis is true. This means that need take an odd value of $m$, say $3$ and need not take any even term as the base case had $2$ terms.

For $m=3$ (i.e., three terms as the simplest case above base), the value of the sum (for any value $p$) is : $1/p + 1/(p+1) + 1/(p+2)$ => $((p+1)(p+2) + (p)(p+2) + (p)(p+1))/(p(p+1)(p+2))$. But, need consider this expression for both $p=2k$ (even) and $p = 2k +1$ (odd), for any integer $k$.

Case (i) $p= 2k : \frac{((2k+1)(2k+2) + (2k)(2k+2) + (2k)(2k+1))}{(2k(2k+1)(2k+2))}$

=> $\frac{(odd.even + even.even + even.odd)}{(even.odd.even)}$

Need consider the fact that the denominator is having an extra multiplier as compared to any term in the numerator. But, to turn this fact into an additional result that simple reduced form, will cancel out the even term in the numerator while still leaving out the denominator as even, is difficult. It also means that need prove that there is an additional even term in the denominator, as compared to the numerator.

While this proving may seem logical in case (i), but in case (ii) where the denominator has only one even term (i.e., $2k+2$), it seems illogical.

Case (ii) $p= 2k+1 : \frac{((2k+2)(2k+3) + (2k+1)(2k+3) + (2k+1)(2k+2))}{((2k+1)(2k+2)(2k+3))}$

=> $\frac{(even.odd + odd.odd + odd.even)}{(odd.even.odd)}$

I hope that this approach, either is done wrong by me, or induction would not work here.

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    $\begingroup$ What happens when n < 0? Is p a prime number? $\endgroup$ – arriopolis Dec 17 '17 at 1:43
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    $\begingroup$ @arriopolis In fact, what happens if $n=0$? $\endgroup$ – Alexander Burstein Dec 17 '17 at 1:45
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    $\begingroup$ @arriopolis Nothing stated like that, the problem is from Uspensky, Heaslet's book on NT. Nothing stated like that in the chapter of which it is an exercise question, regarding $p$ being a prime. $\endgroup$ – jiten Dec 17 '17 at 1:50
  • $\begingroup$ @arriopolis Sorry for late editing, although nothing stated in the book of that sort, I have assumed it to be for Naturals. $\endgroup$ – jiten Dec 17 '17 at 2:11
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Let us assume $n\geq 1$.

If $2^m$ ($m\geq 1$) is the largest power of $2$ which divides some element among $p,p+1,\ldots,p+n$, it divides exactly one element. By letting $M=\text{lcm}(p,p+1,p+2,\ldots,p+n) = 2^m\cdot D$ we have that

$$ 2^{m-1}D\sum_{k=0}^{n}\frac{1}{p+k}\in\mathbb{Z}+\frac{1}{2}$$ and the claim simply follows.

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  • $\begingroup$ Can you please elaborate the answer, with some example. $\endgroup$ – jiten Dec 17 '17 at 1:55
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    $\begingroup$ @jiten: consider, for instance, $p=5$ and $n=6$. The largest power of $2$ dividing some element of $\{5,6,7,8,9,10,11\}$ is $2^3$, and the only element of $\{5,6,7,8,9,10,11\}$ which is a multiple of $2^3$ is $8$. It follows that $\frac{1}{5}+\frac{1}{6}+\ldots+\frac{1}{11}$ multiplied by $5\cdot 3\cdot 7\cdot 9\cdot 5\cdot 11$ is some integer multiple of $\frac{1}{4}$ plus $\frac{1}{8}$, i.e. a number with an even denominator. $\endgroup$ – Jack D'Aurizio Dec 17 '17 at 1:59
  • $\begingroup$ Thanks a lot for giving such a fitting approach, but still unable to understand as to how it would be applicable. I tried a simpler example, but unable to complete it due to even terms, and am confused. Rather than taking it long, I would request you to complete the below : $p=2, n=2 , \frac{1}{2} + \frac{1}{3} + \frac{1}{4}$. The largest power of $2$ is $2^2=4$. It follows that $\frac{13}{12}$ multiplied by ... $\endgroup$ – jiten Dec 17 '17 at 2:43
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    $\begingroup$ @jiten: $3$ has an even denominator. $\endgroup$ – Jack D'Aurizio Dec 17 '17 at 2:44
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    $\begingroup$ @jiten: both approaches work and they are equivalent to a classical argument for showing (through $\nu_2$) that $H_n\not\in\mathbb{Z}$ for any $n>1$. Have a look at dropbox.com/s/auxzc1w0mubpx55/… $\endgroup$ – Jack D'Aurizio Dec 17 '17 at 3:11

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