The total energy of an electrostatic system My problem is from Griffiths Introduction to Electrodynamics, Fourth Edition, p.112 Problem 2.60 (not homework):

A point charge $q$ is at the center of an uncharged spherical
  conducting shell, of inner radius $a$ and outer radius $b$. Question:
  How much work would it take to move the charge out to infinity
  (through a tiny hole drilled in the shell)? [Answer:
  $(q^2/4\pi\epsilon_0)(1/a)$.]

So we basically need to calculate the total energy of the system (excluding the self-energy of the point charge). I did it, but my result deviated from the answer provided. Below is my solution.
I tried to calculate the energy with the formula
$$W = \frac{1}{2} \sum_{i=1}^n q_i V(\textbf{r}_i),$$
or
$$W = \frac{1}{2} \int \rho V d \tau.$$
Here we have three "objects": the point charge $q$ (I will use subscript $\text{c}$ for it), the inner shell with charge $-q$ uniformly distributed, and the outer shell with charge $q$ uniformly distributed. Taking into account the superposition principle of potentials, we have
$$
\begin{aligned}
V_\text{c} 
&= V_{\text{inner}\to\text{c}} + V_{\text{outer}\to\text{c}} \\
&= \frac{-q}{4\pi\epsilon_0 a} + \frac{q}{4\pi\epsilon_0 b},
\end{aligned}
$$
here the self-action is avoided, and
$$
\begin{aligned}
V_\text{inner} 
&= V_{\text{c}\to\text{inner}} + V_{\text{inner}\to\text{inner}} + V_{\text{outer}\to\text{inner}}\\
&= \frac{q}{4\pi\epsilon_0 a} + \frac{-q}{4\pi\epsilon_0 a} + \frac{q}{4\pi\epsilon_0 b}\\
&= \frac{q}{4\pi\epsilon_0 b},
\end{aligned}
$$
$$
\begin{aligned}
V_\text{outer} 
&= V_{\text{c}\to\text{outer}} + V_{\text{inner}\to\text{outer}} + V_{\text{outer}\to\text{outer}}\\
&= \frac{q}{4\pi\epsilon_0 b} + \frac{-q}{4\pi\epsilon_0 b} + \frac{q}{4\pi\epsilon_0 b}\\
&= \frac{q}{4\pi\epsilon_0 b}.
\end{aligned}
$$
Therefore, the total energy is
$$
\begin{aligned}
W 
&= \frac{1}{2}(q V_\text{c} + (-q) V_\text{inner} + q V_\text{outer})\\
&= \frac{q^2}{8\pi\epsilon_0}\left(\frac{1}{b} - \frac{1}{a}\right).
\end{aligned}
$$
What's wrong with my argument? And what is the right way to do the problem?
 A: Your calculation is nice, and the answer seems correct. Another way (less nice) to calculate it is to imagine the charge in the center as a small charged ball (to avoid infinities) and calculate by how much is the field energy lower when it is in the center than when it is far away. Due to symmetry and the Gauss law, the electric field in the former case is the same as if the charge was free, except for the space within the metal. The missing energy is
$$
\Delta U = \int_a^b \frac{1}{2}\epsilon_0 E(r)^2 4\pi r^2 dr,
$$
where $E(r)$ is the electric field of the center charge in the space of the metal, that is, $E(r)  =\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}$. The integral gives the value
$$
\Delta U = \frac{Q^2}{8\pi\epsilon_0}\left(\frac{1}{a}-\frac{1}{b}\right).
$$
As the separated case has energy 0, the case with charge in the center has energy given by minus the above value, which is the same you found above.
A: Your answer is correct, except that it should be negative, i.e.
$$W=\frac{q^2}{8\pi \epsilon_0}\left(\frac{1}{a}-\frac{1}{b}\right)$$
Actually, this is a mistake in the first printing of Griffiths' textbook. You can read the errata here:
http://academic.reed.edu/physics/faculty/griffiths/ED4a.pdf
