Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Assume $k,n$ are positive integers. The sum

$\dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+\ldots +\dfrac{1}{k+n-1}+\dfrac{1}{k+n}$

is not an integer.

The case $k=0$ is proved in this question "Is there an elementary proof that $\sum_{k=1}^{n}1/k$ is never an integer?"

share|improve this question
    
Note that your sum is merely $H_{k+n}-H_k$; a proof similar to the one in the linked question might work here as well. –  J. M. Sep 22 '10 at 15:32

2 Answers 2

up vote 9 down vote accepted

The proof that I gave in that thread works just as well here. It depends only on the fact that in any contiguous sequence of integers (here denominators) the maximal power $\rm 2^k$ that divides any element occurs in precisely one element. Indeed, after the first (necessarily odd) multiple of $\rm 2^k$, the next multiple would, by contiguity, be an even multiple, so a multiple of $\rm 2^{k+1}$, contra maximality. Here is said proof:

HINT $\;$ Since there is a unique denominator $\rm 2^k$ having maximal power of $2$, upon multiplying all terms through by $\rm 2^{k-1}$ one deduces the contradiction that $\rm\; a/2 = b/c \;$ with $\rm \; a,\ c \:$ odd. As an example:

$\quad\quad\quad\quad\quad\quad m = \frac{1}{2} + \frac{1}{3} +\; \frac{1}{4} \;+\; \frac{1}{5} + \frac{1}{6} + \frac{1}{7} $

$\quad\quad\Rightarrow\quad\;\; 2m = \; 1 + \frac{2}{3} +\; \frac{1}{2} \;+\; \frac{2}{5} + \frac{1}{3} + \frac{2}{7} $

$\quad\quad\Rightarrow\quad -\frac{1}{2} = \; 1 + \frac{2}{3} - 2m + \frac{2}{5} + \frac{1}{3} + \frac{2}{7}$

The prior sum has all odd denominators so reduces to a fraction with odd denominator $\rm d\:|3\cdot 5\cdot 7$.

Note $\:$ I purposely avoided any use of valuation theory because Anton requested an "elementary" solution. The above proof can easily be made comprehensible to a high-school student. The proof is trivial to anyone who knows valuation theory: namely the sum has a lone dominant term with 2-adic value $\rm v_2<0\:$ so, by the domination principle, the sum has 2-adic value $\rm v_2<0\:,\:$ i.e. the sum has even denominator in lowest terms, so it is nonintegral.

share|improve this answer
    
Would you mind transcribing here your proof? –  Américo Tavares Sep 22 '10 at 16:17

{Differences of harmonic numbers are not integers}

Consider $\sum_{k=m}^n {1\over k}$. Find the largest power of $2^r$ that divides one of the denominators between $m$ and $n$. There can be no even multiples of $2^r$ between $m$ and $n$, hence there is only one odd multiple.

Therefore $2^{r-1}(1/m+\cdots+1/n)$ is equal to $1/2o_1$ plus a bunch of fractions with odd denominator. Adding them gives one fraction with an odd denominator, write it as $x/o_2$. Here $o_1$ and $o_2$ are odd integers.

We have $2^{r-1}(1/m+\cdots+1/n)=1/2o_1+x/o_2$ so that $${1\over m}+\cdots+{1\over n}={2xo_1+o_2\over 2^r o_1 o_2}.$$ Therefore $2^r$ divides the denominator. If $n\not=m$, then $r\geq1$ and the sum is not an integer.

Cf. Exercise 6.21 (page 311) of Graham, Knuth, and Patashnik.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.