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Define the sequence $r(k)= \sum_{n=2}^k 2^{\frac{1}{n}}$

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

3 Answers 3

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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 $r(k)\not\in\mathbb{Q}$.

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    Are you implicitly using Bertrand's postulate here? – 2017-02-11
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    Would you mind elaborating why the sum has degree $pm\geq p$? Reasoning of this kind seems to apply equally well to $\sqrt{2}+\sqrt[3]{2}+(-\sqrt{2}-\sqrt[3]{2})$ - there is precisely one term of degree $3$, but the sum is not of degree divisible by $3$. – 2017-02-11
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    @HagenvonEitzen: yes I am. Is that an issue? The first cases ($k\leq 6$) can be checked by hand. – 2017-02-11
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    @Wojowu You skipped a sole term of degree $5$ that that is present in the sum in the question. – 2017-02-11
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    @Arthur I am aware of that. But why is it relevant? – 2017-02-11
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    @Wojowu: if $[\mathbb{Q}(\alpha):\mathbb{Q}]=p$ and $[\mathbb{Q}(\beta):\mathbb{Q}]=q$ with $\gcd(p,q)=1$, then the degree of $\alpha+\beta$ is simply $pq$. – 2017-02-11
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    Ah, so you are actually using that there is only one term with degree divisible by $p$. I see how that helps, thanks. – 2017-02-11
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    @JackD'Aurizio No, it's not. I just think it is worth mentioning. – 2017-02-11
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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).

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    The same argument shows that if $a_k,b_k\in\mathbb Q$ and $\sum a_k^{b_k}\in\mathbb Q$ with each term irrational, then $\sum a_k^{b_k}=0$. Nice! – 2017-02-11
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    If you assume $ a_k > 0 $ for all $ a_k $ and take the real roots, then yes, this holds. (To see why this is necessary, consider that the sum of the conjugates of $ \zeta_3 $ is $ -1 $, for instance...) – 2017-02-11
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    Oops, indeed. Or more generally if $|a_k^{b_k}|\notin\mathbb Q$, with arbitrary choices of the roots. – 2017-02-11
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    @barto That doesn't work, because the trace of $ (-4)^{1/4} $ is either $ 2 $ or $ -2 $, depending on your choice of complex root. – 2017-02-12
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    Thanks, I was assuming that if $|a^{1/n}|\notin\mathbb Q$ every irreducible factor of $x^n-a$ of degree $d$ has no term of degree $d-1$ (which may indeed fail if e.g. $a$ is not square-free). So much for my attempts to formulate a generalization... – 2017-02-12
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    Pardon my naivety, but why is it clear the trace kills it? – 2017-02-12
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    It follows from the transitivity of the trace, and the fact that $ X^n - 2 $ is irreducible, so that the trace of $ 2^{1/n} $ in $ \mathbf Q(2^{1/n}) $ is $ 0 $. – 2017-02-12
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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