TITLE: Given $1 + 2^{1/2} + 3^{1/3} + \dots + 100^{1/100}$ lies strictly between 111 and 112, show it is irrational
QUESTION [7 upvotes]: My title completely describes the question. I tried to look at the polynomial $$p(x) = (x-1)(x-2^{1/2})\dots (x-100^{1/100})$$ since the sum naturally appears as the coefficient of $x^{99}$, and then tried examining $p$ at a few values like $1, 99 \ \& 100$, but didn't seem to be useful. My guess is that the numbers $111$ and $112$ don't really matter but the fact that the sum is not an integer may be essential. Any hints?

REPLY [8 votes]: Following Empy2's comment, here is a solution. Since $n^{1/n}$ is a root of $x^n - n$, $n^{1/n}$ is an algebraic integer for $n = 1,2,\dots,100$. Since the algebraic integers form a ring, their sum is also an algebraic integer. Because we know from the rational roots theorem that $\mathbb{Z}_\mathbb{Q} = \mathbb{Z}$, and since the real number $1 + 2^{1/2} + \dots + 100^{1/100}$ is not an integer, it must be an irrational number.