FTSOC suppose this never holds. Then note that $2 > x_i > 1$ for all $i$. Furthermore, if $x_n < 1 + \frac{1}{n}$ then $x_{n+1} = x_n$, else $x_{n+1} = x_n \cdot \frac{n}{n+1}$. This means we can instead the sequence \[ y_{n+1} = y_n \cdot \frac{k}{k+1} \]where $k$ is minimal such that $y_{n} > \frac{k+1}{k}$. If equality ever holds, then we get a contradiction as that would imply $x_i = 1$. However, notice that if $y = \frac{p}{q}$ and $k$ is taken to be minimal then \[ \frac{p}{q} < \frac{k}{k-1} \iff p(k-1) < qk \iff pk - q(k+1) < p - q \]so the monovariant $p_n - q_n$ for $y_n = \frac{p_n}{q_n}$ always decreases.

Posted a wrong solution, fixed it, only to another sol be posted and spend more time trying to make the lim x_n = 1 being useful Call an index $i$ from the sequence good if $x_i < \frac{i+1}{i}$, and extraordinary otherwise. The sequence is then not altered in any good index, but only in extraordinary ones. Let $i$ be good with $i+1$ extraordinary, then \[ 1 + \frac{1}{i+1} \le x_{i+1} = x_i < 1 + \frac{1}{i} < 1 + \frac{2}{i+1} \]If we only have a good numbers after a certain point, then $x_i = 1$ for some $i$, otherwise, if equality ever holds in extraordinary numbers, we are also done. The main claim is now that for the quantity $p_i - q_i$ will converge $0$ when $x_i = \frac{p_i}{q_i}$, we have that for an extraordinary $i$, then \[\frac{p_{i+1}}{q_{i+1}} = \frac{p_i}{q_i}\frac{i}{i+1}\]So, \[p_{i+1} - q_{i+1} \le p_{i}i - q_{i}(i+1) = p_{i}(i-1) - q_{i}i + p_i - q_i < p_i - q_i\]Because $p_i$ and $q_i$ have bounded difference and are integers, the convergence will imply that the sequence is $0$, and we are done! $\Box$

note to self

Cool opening problem which is not essentially easy but is pretty accessible. I hope this solution works. We first show that the exists some term $1<x_k<2$ in this sequence. This is quite straightforward since if $x_1<2$ we are already done and otherwise, letting $x_1 = \frac{kq+r}{q}$ for positive integers $k,q$ and $r$ we have, \begin{align*} x_2 &= \frac{2x_1}{\lfloor 2x_1 \rfloor}\\ &= \frac{\frac{2kq+2r}{q}}{\lfloor \frac{2kq+2r}{q} \rfloor}\\ \end{align*}Now, if $2r<q$ we have, \[x_2 = \frac{\frac{2kq+2r}{q}}{2k}= \frac{2kq+2r}{2kq} < \frac{2kq+q}{2kq} < 2\]and if $2r>q$ (but $2q>2r$) we have \[x_2 = \frac{\frac{2kq+2r}{q}}{2k+1}=\frac{2kq+2r}{2kq+q} < \frac{2kq+2q}{2kq+q}<2\]Thus, $1\le x_2<2$, and such a term does indeed always exist. Note that our sequence $(x_i)$ is non-increasing. This is simply because if there exists $x_n>1$ such that $x_{n+1}=\frac{nx_n}{\lfloor nx_n \rfloor}>x_n$ then, \[n > \lfloor nx_n \rfloor > nx_n -1 > n-1\]which is a clear contradiction since then $\lfloor nx_n \rfloor$ is a positive integer between $n$ and $n-1$ (which simply does not exist). Next, our key claim is the following. Claim : For all $i\ge 2$ (for which $x_i <2$), let $x_i = \frac{q+r}{q}$ for positive integers $q$ and $r$. Define the sequence $(a_n)_{n\ge 2}$ such that for all $i\ge 2$ , $a_i=r$. Then, $(a_n)$ is monotonically decreasing (and must decrease at certain points). Proof : Consider $x_i = \frac{q+r}{q}$ for some $i\ge 2$. Then for all $a>i$, such that $ar<q$ we have, \[x_{a+1} = \frac{\frac{q(q+r)}{q}}{\lfloor \frac{q(q+r)}{q} \rfloor} = \frac{\frac{a(q+r)}{q}}{a} = \frac{q+r}{r}=x_a\]Thus, the sequence $(a_n)$ stays constant in this range. Now, let $b$ be the minimum positive integer such that $br\ge q$. Then, \[x_{b+1} = \frac{bx_b}{\lfloor bx_b \rfloor} = \frac{\frac{b(q+r)}{q}}{q(b+1)}= \frac{bq+br}{bq+q}\]Then, \begin{align*} a_{b+1} &= (bq+br) - (bq+q)\\ &= br-q\\ &= (b-1)2 - q + r\\ & < q-q +r\\ &= r\\ &= a_b \end{align*}So, $(a_n)$ has decreased. Thus, this sequence is non-increasing as claimed, and must decrease at certain terms. Now, this means that eventually there exists a term $a_k=0$. But then, \[x_k = \frac{n+a_k}{n} = 1\]for some positive integer $n$. Thus, $x_k=1$ for this choice of $k$, and a term of the type we wished indeed exists and we are done.

İs this enough? Assume that $x_i>1$ for all $i$. Then obviously $x_{n+1} \le x_n$. So the sequence should become eventually constant say $c$. Now for every $n \ge N$, we should have $c=\frac{c \cdot n}{\lfloor c \cdot n \rfloor}$ so $c<\frac{n+1}{n}$ for every $n \ge N$.Obviously $c \le 1$ contradiction.