Just choose $\varepsilon_{n}$ for $n\in\mathbb{N}$, so that $\sum_{i=1}^{n}\varepsilon_{i}=\lfloor\alpha n\rfloor$ for all $n\in\mathbb{N}$. Now let us prove this works. We will say $(a_i)_{i\geq1}\geq(b_i)_{i\geq1}$ if $\sum_{i=1}^{n}a_i\geq\sum_{i=1}^{n}b_i$ for all natural $n$. It is easy to prove that if $(a_i)_{i\geq1}\geq(b_i)_{i\geq1}$ then $\sum_{i=1}^{n}\frac{a_{i}}{(n+i-1)(n+i)}\geq\sum_{i=1}^{n}\frac{b_{i}}{(n+i-1)(n+i)}$ for all natural $n$. Now let $(\delta_i)_{i\geq1}$ be chosen so that $\sum_{i=1}^{n}\delta_{i}=\lfloor\alpha n\rfloor+1$ for all $n$. Then $(\delta_i)_{i\geq1}\geq(\alpha)_{i\geq1}\geq(\varepsilon_i)_{i\geq1}$. Now notice that $2n\sum_{i=1}^{n}\frac{\alpha}{(n+i-1)(n+i)}=2n\sum_{i=1}^{n}\alpha(\frac{1}{n+i-1}-\frac{1}{n+i})=2n\alpha(\frac{1}{n}-\frac{1}{2n})=\alpha$. So $\alpha-2ns_n\geq0$. Now it's enough to prove that $2n\sum_{i=1}^{n}\frac{\delta_i}{(n+i-1)(n+i)} - 2n\sum_{i=1}^{n}\frac{\varepsilon_i}{(n+i-1)(n+i)}=\frac{2}{n+1}$. Notice that $\varepsilon_1=0$, $\delta_1=1$ and $\varepsilon_{i}=\delta_{i}$ for $i\geq2$. Then this difference is equal to $2n\frac{1}{n(n+1)}=\frac{2}{n+1}$, done.

Very nice

this problem can also use Abel to solve.