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Assume that $n>1$ is a positive integer. I am trying to prove that $$S=1+{\sqrt{2}}+{\sqrt[3]{3}}+\cdots+{\sqrt[n]{n}}$$ must be an irrational number. I seek a proof that is elementary and does not use algebraic field theory, or an explanation why a proof using field theory will not work. Some commentors on an earlier version of this post have provided very useful links to proofs involving field theory, already posted on stack exchange, and references to papers that I hope to work through. Another commentor seemed to believe that my idea for a proof would not work-which may well be the case! (Please see the useful comments by Martin Brandenburg and R.P. below). It would be helpful to know why this is so. I am merely stating the idea that occurred to me.

My approach is as follows. Assume that $[S]$ is the integral part of the above sum $S$. I hope to prove that $$[S]<S<[S]+1.$$ In particular, that both displayed inequalities are strict. If I can prove that this is the case, I would be done. This is because in the first place, $S$ must be an algebraic integer. This follows since the set of algebraic integers form a subring of the field of algebraic numbers. And since every summand ${\sqrt[k]{k}}$ (being a root of $x^{k}-k$) is an algebraic integer, $S$ being a finite sium of algebraic integers must also be one. Now, if $S$ happens to be a non-integral rational number, and an algebraic integer, then $S$ must be an ordinary integer, a conclusion that is precluded by the strict inequalities displayed above.

student
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    Almost-Duplicate of https://math.stackexchange.com/questions/890821/proof-that-sqrtma-sqrtnb-is-irrational (see the answer and apply the method). – Martin Brandenburg Dec 23 '24 at 04:05
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    This question is similar to: Sum of irrational numbers, a basic algebra problem. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. FYI, this was found linked to in Martin's almost-duplicate mentioned above. Also, there's $\sqrt[b_1]{a_1}+\sqrt[b_2]{a_2}+\cdots+\sqrt[b_k]{a_k}$ not an integer, in the right side related section, which can fairly easily be used. ... – John Omielan Dec 23 '24 at 04:12
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    (cont.) In particular, assume using this result that instead of an integer the sum is a rational non-integer, say $\frac{c}{d}$ where $c$ and $d$ are positive integers with $\gcd(c,d)=1$. Then multiply both sides by $d$, change each left side term to use that $d\sqrt[b_i]{a_i}=\sqrt[b_i]{d^{b_i}a_i}$ is also not an integer, and finally apply the question's result to this new summation to get a contradiction. – John Omielan Dec 23 '24 at 04:18
  • The approach in the links is via field theory, which is illuminating-so happy to get useful insights, but the approach sketched in the question I posted is a little different-estimating the integer part of the sum. Happy to withdraw it if you feel it is a duplicate. – student Dec 23 '24 at 04:52
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    @student You can easily make it a non-duplicate by specifying that you are looking for a solution without field theory (if I understand this correctly) and also only looking for that specific sum (the linked questions are more general). You can edit the question accordingly if you are not satisfied with the linked questions and explain why. – Martin Brandenburg Dec 23 '24 at 11:11
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    It is not hard to show that $n^{1/n} - 1 > \frac{1}{n}$ (if I had to guess I'd say the fractional part of $n^{1/n}$ is something like $\frac{\log(n)}{n}$, but I don't have a proof of this nor a more precise statement), so at least we can be sure that the fractional part of $S$ never stabilizes. To me this raises the question why we would even suspect a proof via bounding $S$ away from an integer is feasible at all. The only reason I have for believing that in fact $S$ never hits an integer is the proof via field theory. – R.P. Dec 23 '24 at 12:36
  • Thanks Martin. I can edit the question. And thanks R.P. Maybe a proof along the lines I outlined is not possible! – student Dec 23 '24 at 15:47
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    I'm not sure how I feel about closing this as duplicate by linking to generalized question - what if this special case can be proven much more simpler? ... – Sil Dec 23 '24 at 16:29
  • A simpler proof would be great! It is a natural sum to consider. I was just playing around.... – student Dec 23 '24 at 16:31
  • R.P. I reasoned that as $$(1+{\frac{1}{n}})^{n} <e<3\leq{n}$$ for $n\geq{3}$, it must be the case that $n^{\frac{1}{n}} -1 >{\frac{1}{n}}$. So these differences sum to infinity (compare with the harmonic series) and are hard to control. Have a sinking feeling you may be right! – student Dec 24 '24 at 20:19
  • Found some information about my question. The minimum polynomial for $S$ was calculated by E. Fried, On linear combinations of roots. Magyar Tud. Akad. Mat. Fiz. Ost. Kozl. 3 (1954) 155-162. The answer is complicated, but the degree is >1, which proves that $S$ must be irrational. The paper was reviewed by Erdos in AMS Math Reviews (MR 0067922 (16 798f) 10.0X). I cannot take credit for my question! It was posed by Sierpinski apparently. An answer involving Diophantine Approximation would be interesting. – student Dec 30 '24 at 17:08

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