Define the sequence $r(k)= \sum_{n=2}^k 2^{\frac{1}{n}}$
Is $r(k)$ irrational for every natural $ k\geq 2$?
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3 Answers
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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 $r(k)\not\in\mathbb{Q}$.
answered Feb 11, 2017 at 22:32
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4$\begingroup$ Are you implicitly using Bertrand's postulate here? $\endgroup$ Commented Feb 11, 2017 at 22:36
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$\begingroup$ 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 $3$. $\endgroup$– WojowuCommented Feb 11, 2017 at 22:37
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1$\begingroup$ @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 $pq$. $\endgroup$ Commented Feb 11, 2017 at 22:40
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2$\begingroup$ Ah, so you are actually using that there is only one term with degree divisible by $p$. I see how that helps, thanks. $\endgroup$– WojowuCommented Feb 11, 2017 at 22:42
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4$\begingroup$ @JackD'Aurizio No, it's not. I just think it is worth mentioning. $\endgroup$ Commented Feb 11, 2017 at 23:08
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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 $\frac k2<p\le k$ (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 $r(k)\notin\Bbb Q$.
answered Feb 11, 2017 at 23:07
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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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$\begingroup$ 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! $\endgroup$ Commented Feb 11, 2017 at 23:23
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$\begingroup$ 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...) $\endgroup$ Commented Feb 11, 2017 at 23:25
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$\begingroup$ Oops, indeed. Or more generally if $|a_k^{b_k}|\notin\mathbb Q$, with arbitrary choices of the roots. $\endgroup$ Commented Feb 11, 2017 at 23:58
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$\begingroup$ @barto That doesn't work, because the trace of $ (-4)^{1/4} $ is either $ 2 $ or $ -2 $, depending on your choice of complex root. $\endgroup$ Commented Feb 12, 2017 at 0:08
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1$\begingroup$ 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 $. $\endgroup$ Commented Feb 12, 2017 at 6:38