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Given the series $$S=\sum_{k=1}^N\frac{1}{k^q}$$ $(q\gt0),(N\in\mathbb{N},N\ge1)$ is $S$ irrational for every choice of $q$ and $N$? Thanks.

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Are you missing something? If $q\in\mathbb{N}$, then $S$ is rational. –  Casteels Jun 13 at 9:42
    
@Casteels: $q\in\mathbb{R}$ –  Riccardo.Alestra Jun 13 at 9:47
    
Do you mean $q\in\mathbb{R}\setminus\mathbb{Z}$? –  Casteels Jun 13 at 9:53
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The question might be better if you asked if there are any non-integer reals $q$ making the sum rational. –  coffeemath Jun 13 at 9:53
    
I fully support coffeemath's suggestion. May be you could rephrase and precise this interesting question. –  Claude Leibovici Jun 13 at 9:55

3 Answers 3

up vote 7 down vote accepted

$S$ is a continuous function of $q$, which is rational for integer values of $q$. So it can't possibly be irrational for all non-integer values of $q$.

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+1 very nice, simple, straighforward yet not trivial argument :) –  tohecz Jun 13 at 22:25

For fixed $N$ the value of $S$ is a continuous function of the real number $q$, so one can just choose some rational number $r$ and then let $S=r$ and there will exist a $q$, provided $r$ is in the right range.

I think you can rescue the result provided one restricts $N \ge 2$ and $q$ to be a positive non-integral rational number. If $q=r/k$ with $k>1$ then for the case $N=2$ one can use that $2^{1/k}$ is irrational, and for larger $N$ then a collection of distinct primes each raised to the power $1/k$ are rationally independent. [That last claim I've seen somewhere but don't have details handy.]

It now seems better to use that $2^{r/k}$ is irrational, and that a collection of distinct primes each raised to the power $r/k$ are rationally independent. The drawback on getting rid of the numerator of the rational $r/k$ is that then one is working with powers of supposed irrationals, which are not immediately seen to be irrational / independent (though they are in this case).

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This does not fully answer the question but it could help (I hope). Since $$S=\sum_{k=1}^n\frac{1}{k^q}=H_n^{(q)}$$ let us consider the case of rational values of $q$ and set $q=\frac{a}{b}$. In such a case, we have the identity $$H_n^{\left(\frac{a}{b}\right)}=\zeta(n)-b^n \sum_{k=1}^\infty \frac {1}{(a+bk)^n}$$ Should we then address the problem of the rationality of function $\zeta(n)$ ?

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