For the summation $\sum_{n=0}^x \sqrt{n}$ are there any values of $x$ where the summation equals a whole number other than 1?

  • $\begingroup$ @Rob Arthan thanks for pointing out . $\endgroup$ – N.S.JOHN Feb 26 '16 at 3:14

I doubt there's an elementary proof of this, but the fact that this is irrational for $x>1$ follows from the following theorem:

The set of numbers of the form $\{\sqrt{n}:n\in \mathbb Z,\,n \text{ squarefree}\}$ is linearly independent over $\mathbb Q$.

One can find proofs for this here and in the special case for primes here.

This is essentially an immediate corollary: Suppose for contradiction that $\sum_{n=1}^x\sqrt{n}=q$ was rational. Every term in the sum may be written of the form $c_n\sqrt{d_n}$ where $d_n$ is square-free and $n=c_n^2d_n$ and both $c_n$ and $d_n$ are integers. If we regroup the terms to collect coefficients of $\sqrt{d_n}$ we'll get $$\sum_{\substack{d=1\\d\text{ squarefree}}}^{x}\left(\sum_{c=1}^{\lfloor\sqrt{x/d}\rfloor}c\right)\sqrt{d}$$ and if we subtract $q$ from this and pull out the first term, we get $$\left(-q+\sum_{c=1}^{\lfloor\sqrt{x}\rfloor}c\right)\cdot \sqrt{1}+\sum_{\substack{d=2\\d\text{ squarefree}}}^{x}\left(\sum_{c=1}^{\lfloor\sqrt{x/d}\rfloor}c\right)\sqrt{d}=0$$ which contradicts that the set of square roots of square free numbers is linearly independent over $\mathbb Q$. Therefore, the sum $q$ is not rational - and, in particular, is not an integer.

One should note that this gives a stronger property: If you sum up a bunch of square roots of natural numbers with positive rational coefficients, where at least one of the square roots is irrational, the sum will never be rational.


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