Is $3^{1/2}+3^{1/3}+3^{1/4}+3^{1/5}+\cdots+3^{1/(n+1)}$ always irrational?

Question

$$3^{1/2}+3^{1/3}+3^{1/4}+3^{1/5}+\cdots+3^{1/(n+1)}$$ where $n \in \mathbb{N}$

Is this always irrational?

If it is always irrational, how can we prove it?

I am second year student of mathematics I know calculus 1+2+3 ,ODES,logic and writing proofs . at my current semester I am studying Abstract Algebra 01 ,Number Theory , Real Analysis ,PDES 01 and Linear Algebra 01. This question comes to my mind while I am thinking about the uncountability of irrationals

Answer 1

Let $N=\operatorname{lcm}(2,3,\ldots,n+1)$. Observe that $N$ is even, and that one of the indices $2,3,\ldots,n+1$, is divisible by the same power of two as $N$. Let $\alpha=3^{1/N}$ be the real and positive $N$th root of $3$. The sum, denote it by $\Sigma$, is an element of the field $K=\Bbb{Q}(\alpha)$.

Roadmap: Describe a non-trivial automorphism $\sigma$ of $K$. Show that $\Sigma$ is not a fixed point of $\sigma$. Conclude.

Answer 2

Here is another proof, which I consider simpler. Following Jyrki Lahtonen we let $N=\mathrm{lcm}(2,3...n+1)$, $\alpha=3^{1/N}$, and $\Sigma$ be the sum in question. Moreover, let

$$\Sigma(x)=\sum_{k=2}^{n+1}x^{N/k}=\mathrm{degree\,\,}N/2\,\,\mathrm{polynomial}$$

Which has $\Sigma(\alpha)=\Sigma$. Thus, the polynomial $\Sigma(x)-\Sigma$ has degree $N/2$ and $\alpha$ is a root, which is a contradiction since we know that $p(x)=x^{N}-3$ is irreducible by Eisenstein's criterion so $\deg(\alpha)=N$. Q.E.D, no automorphisms needed.