 Nonzero Ext of the rationals with integer coefficients

A free resolution of $\mathbb{Q}$ has generators $a_n=1/n!$ and independent relations $a_n-(n+1)a_{n+1}$. Thus the Ext functor gives the cokernel of the map on integer sequence productsThe constant sequence one is not in its image. A preimage would require $u_1\equiv {\sum }_{j=1}^{N}j!(\mathrm{mod}(N+1)!)$ for every $N$. For $N\ge 3$, the sum is below $2N!$, tends to infinity, and its distance from $(N+1)!$ also tends to infinity. No fixed integer satisfies all these congruences. Hence ${\mathrm{Ext}}_{\mathbb{Z}}^1(\mathbb{Q},\mathbb{Z})\ne 0$.