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None of the Harmonic numbers $H_n = \sum_{k=1}^n 1/k$ are integers for $n>1$ (e.g., this MSE question and answer).

Q. Define the $r$-th root Harmonic number $H_n^{1/r} = \sum_{k=1}^n 1/{k^{1/r}}$. E.g., for $r=2$, $H_n^{1/2} = \sum_{k=1}^n 1/\sqrt{k}$. Is $H_n^{1/r}$ ever an integer for $n>1$ and $r \ge 2$ an integer?

Of course there are many close calls, e.g., $H_{202}^{1/2} \approx 27.0002$, $H_{1132}^{1/3} \approx 162.00004$, $H_{222}^{1/4} \approx 76.000003$.

Answered in comments by Vesselin Dimitrov: No. In fact all these sums are irrational.

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    $\begingroup$ They are always irrational. Quite generally, you may show that for any positive rationals $a_i \in \mathbb{Q}^{> 0}$ and any positive integers $k_i \in \mathbb{N}$, a finite sum $\sum_1^n \sqrt[k_i]{a_i}$ is always irrational unless it is rational term-by-term. $\endgroup$
    – Vesselin Dimitrov
    Commented Aug 9, 2014 at 1:56
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    $\begingroup$ We may also consider real irrational exponents $\lambda = 1/r \in \mathbb{R}$. For a given $\lambda \in \mathbb{R} \setminus \mathbb{Q}$, are there only finitely many $n$ for which $H_n^{\lambda}$ is an integer? (a rational?) Is there a uniform bound on the number of exceptions? Surely the answer is "yes," the point is whether it is possible to prove this or not. It was a Putnam contest problem that unless $\lambda$ is a (non-positive) integer, not all $H_n^{\lambda}$ can be integers. $\endgroup$
    – Vesselin Dimitrov
    Commented Aug 9, 2014 at 2:56

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