Proof of irrationality of a series 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.
 A: $S$ 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 $q$.
A: 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).
A: 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 $\zeta(n)$ ?
