= Solution
Use zero-based labels to agree with $\phi_0$ in the later parts. The normalized <infinite square well> <energy eigenstates> and <energy eigenvalues> are
$$
\phi_n(x)=\frac1{\sqrt a}\sin\left[\frac{(n+1)\pi(x+a)}{2a}\right],
\qquad \epsilon_n^{(0)}=K(n+1)^2,\quad
K=\frac{\pi^2\hbar^2}{8ma^2},\quad n=0,1,\ldots.
$$
The <position matrix elements in a symmetric infinite square well>, obtained by integrating the product of the two <wavefunctions> against $x$, are
$$
x_{kn}=\langle\phi_k|x|\phi_n\rangle
=\frac{8a(k+1)(n+1)[(-1)^{k+n}-1]}
{\pi^2[(k+1)^2-(n+1)^2]^2}\quad(k\ne n),\qquad x_{nn}=0.
$$
The integral in the question uses positive integer labels; replacing them by $k+1,n+1$ is essential here. Only opposite-parity states, $k+n$ odd, contribute, as also follows from the <parity operator> <selection rule>.
<Second-order nondegenerate perturbation theory> now gives the requested infinite sum:
$$
\boxed{\delta\epsilon_n^{(2)}=e^2E^2\sum_{k\ne n}
\frac{|x_{kn}|^2}{\epsilon_n^{(0)}-\epsilon_k^{(0)}}.}
$$
In fully explicit form this is
$$
\boxed{\delta\epsilon_n^{(2)}=
-\frac{2048me^2E^2a^4}{\pi^6\hbar^2}(n+1)^2
\sum_{\substack{k\geq0\\k+n\ \mathrm{odd}}}
\frac{(k+1)^2}{[(k+1)^2-(n+1)^2]^5}.}
$$
The symbol $m$ here denotes the electron mass, not a summation index. For the <ground state> $n=0$, every denominator in the first form is negative, so the ground <energy level> moves downward. For excited states, contributions from lower and higher <energy levels> have opposite signs.
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