Is sum of roots of 2 always irrational? Define the sequence $r(k)= \sum_{n=2}^k 2^{\frac{1}{n}}$

Is $r(k)$ irrational for every natural $ k\geq 2$?

 A: Yes. Let we assume $k>4$ and let $p$ the greatest prime $\leq k$. Then $2^{\frac{1}{p}}$ is an algebraic number of degree $p$ over $\mathbb{Q}$ and is the only term of the sum with such a property. It follows that $r(k)$ is an algebraic number over $\mathbb{Q}$ with degree $pm\geq p$ and in particular $r(k)\not\in\mathbb{Q}$.
A: Yes.
The number $r(k)$ is an element of the splitting field $F$ of the irreducible polynomial $f(X)=X^{N}-2$ where $N=\operatorname{lcm}(1,2,\ldots,k)$).
Indeed, if $\alpha$ is the unique positive root of $f$, then we have
$$ r(k)=\sum_{n=2}^k\alpha^{N/n}$$
Let $p$ be a prime with $\frac k2<p\le k$ (such a prime exists by Bertrand's postulate), and let $\beta=\alpha\cdot e^{2\pi i/p}$. Then $p\mid  N$ and so $f(\beta)=0$. Hence there exists an automorphism $\phi$ of $F$ that maps $\alpha\mapsto \beta$. Note that $\beta^{N/n}=\alpha^{N/n}$ for most $n$. Indeed, we have  $\beta^{N/n}\ne\alpha^{N/n}$ if and only if $N/n$ is not a multiple of $p$. As $p^2\nmid N$, this is the case precisely for the case $n=p$. It follows that $$\phi(r(k))=r(k)+\beta^{N/p}-\alpha^{N/p}\ne r(k)$$
and therefore $r(k)\notin\Bbb Q$.
A: Yes, because the field trace of $ \mathbf Q(2^{1/2}, 2^{1/3}, \ldots, 2^{1/n})/\mathbf Q $ kills this number, and the only rational number which is killed by the field trace is $ 0 $ (and the number in the OP is clearly nonzero).
