Let $\{x\}$ denote the fractional part of $x$ and $d_n=\text{lcm}(1,2,...,n)$. Does there exist a sequence with all positive terms $(a_n)_{n\geq 1}$ such that $$\lim_{n\to\infty} \{a_n\}=1\ \text{,}\ \lim_{n\to\infty} \left\{\frac{a_n}{d_n}\right\}=1\ \ \text{and}\ \ \lim_{n\to\infty} \left\{\frac{a_n}{d_n^2}\right\}=\frac{3}{5}$$
Since we have $$\lim_{n\to\infty} \{a_n\}=1$$ Hence $$a_n=b_n-\varepsilon_n \tag{1}$$ where $b_n\in\mathbb{Z}$ and $\varepsilon_n\to 0$ as $n\to\infty$. Since $$ \lim_{n\to\infty} \left\{\frac{a_n}{d_n}\right\}=1$$Hence we have $$\frac{a_n}{d_n}=c_n-\delta_n \tag{2}$$ where $c_n\in\mathbb{Z}$ and $\delta_n\to 0$ as $n\to\infty$. Similarly$$\lim_{n\to\infty} \left\{\frac{a_n}{d_n^2}\right\}=\frac{3}{5}$$This gives $$\frac{a_n}{d_n^2}=p_n+\frac{3}{5}-\eta_n \tag{3}$$ where $p_n\in\mathbb{Z}$ and $\eta_n\to 0$ as $n\to\infty$. Now $(1)$ and $(2)$ gives$$ \frac{b_n}{d_n}-\frac{\varepsilon_n}{d_n}=c_n-\delta_n \tag{4}$$Similarly $(1)$ and $(3)$ gives$$ \frac{b_n}{d_n^2}-\frac{\varepsilon_n}{d_n^2}=p_n+\frac{3}{5}-\eta_n \tag{5}$$Similarly $(2)$ and $(3)$ gives$$ \frac{c_n}{d_n}-\frac{\delta_n}{d_n}=p_n+\frac{3}{5}-\eta_n \tag{6}$$
Any help will be highly appreciated. Thank you!