= Solution
The midpoint-odd perturbation couples the <ground state> only to even-index states. Direct integration gives
$$
\langle1|x-a/2|2j\rangle
=-\frac{16aj}{\pi^2(4j^2-1)^2},\qquad
E_{2j}^{(0)}-E_1^{(0)}
=\frac{h^2}{8ma^2}(4j^2-1).
$$
Insertion in the second-order expression yields
$$
E_1^{(2)}=
-\frac{2048ma^4}{h^2\pi^4}
\sum_{j=1}^\infty\frac{j^2}{(4j^2-1)^5}.
$$
It is negative because all coupled states lie above the <ground state>. This is the coefficient of $\lambda^2$; the physical energy shift includes the additional factor $q^2\varepsilon^2$.
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