Can $\sum_{n=1}^{k} \frac1{\sqrt{n}}$ be an integer for $k>1$? In one of new YouTube videos, I've seen the problem: to show that
$$16 < \sum_{n=1}^{80} \frac1{\sqrt{n}} < 17$$
I've solved the problem easily, using
$$\int_2^{81}\frac1{\sqrt{x}}\, dx < \sum_{n=2}^{80} \frac1{\sqrt{n}} < \int_1^{81}\frac1{\sqrt{x}}\, dx$$
But this problem aimed me on another problem:

Can the sum $\sum_{n=1}^{k}  \frac1{\sqrt{n}}$ be an integer number at some positive integer $k>1$?

Is this problem well-known open or closed problem? Or maybe there is some approach allowing to solve this problem easily without using hard mathematical skills?
I've seen the same problem without square roots and it was solved easily.
 A: This is not an integer due to the same reason that $H_n$ is not an integer.
Take the largest integer $k$ such that $2^k < n$. If $M = 2^{k - 1} \cdot (2n - 1)!!$, then $\frac{\sqrt{M}}{\sqrt{i}}$ is an algebraic integer for any $i \neq 2^k$ between $1$ and $n$, but $\frac{\sqrt{M}}{\sqrt{2^k}}$ is not algebraic integer. So
$$\sqrt{M} \cdot \sum_{i = 1}^n \frac{1}{\sqrt{i}}$$
is not algebraic integer.
A: A linear transformation of a finite dimensional vector space over a field $k$ has a trace.
Now let $k\subset K$ be a finite extension of fields. For $\alpha$ an element of $K$ we have the trace
$$\operatorname{Tr}_{K/k}(\alpha)$$ defined as the trace of the linear operator $x \mapsto \alpha \cdot x$ ( multiplication by $\alpha$).
Fact: if $d$ rational is not a perfect square and $K$ is a field containing $\sqrt{d}$ then $$\operatorname{Tr}_{K/\mathbb{Q}}(\sqrt{d})= 0$$
Indeed: let $\beta_1$, $\ldots$, $\beta_m$ be a basis of $K$ over the subfield $\mathbb{Q}(\sqrt{d})$. Then $\beta_1$, $\ldots$, $\beta_m$, $\sqrt{d} \beta_1$, $\ldots$, $\sqrt{d} \beta_m$ form a basis of $K$ over $\mathbb{Q}$. In this basis the operator "multiplication by $\sqrt{d}$ has a matrix with diagonal elements all $0$. Done.
Now to show that $\alpha \colon = \sum_{d \in M} \sqrt{d}$ is not rational if $M$ is a finite set of positive rationals that are not square. Indeed, consider a finite extension $K$ of $\mathbb{Q}$   containing all these roots. We have $$
\operatorname{Tr}_{K/\mathbb{Q}}(\alpha) = \sum_{d \in M} \operatorname{Tr}_{K/\mathbb{Q}} (\sqrt{d})= 0$$
But the trace of a  rational number $r$ equals $\operatorname{Tr}_{K/\mathbb{Q}}(r)= n r$, where $n = [K\colon \mathbb{Q}]$.   Hence the above positive sum cannot be a rational number.
With a bit more care, we can show that incomensurable roots of rational numbers are linearly independent over $\mathbb{Q}$.
$\bf{Added:}$ Indeed, say we have
$$\sum c_m \rho_m= 0$$
$c_m$ rationals, and  $\rho_m$ are radicals of rationals, and incomensurable ( that is $\frac{\rho_m}{\rho_{m'}} \not \in \mathbb{Q}$). Then we get
$$c_1 = - \sum_{m> 1} c_m \frac{\rho_m}{\rho_1}$$
and taking traces we get $c_1 = 0$. (Note that $\frac{\rho_m}{\rho_1}$ is an irrational radical, so of trace $0$).
