Calculating potential energy per ion of an one dimensional ionic crystal The problem states to calculate "the potential energy, per ion, for an infinite 1D ionic crystal with separation $a$"; the crystal is a 1D lattice of alternating charges, likes so:
$$...+-+-+-+-+-...$$
I believe I have the answer(s) that arrived through two methods; the thing that is vexing me is that the answers are not consistent.
Method 1
Let an arbitrary positive charged ion be the charge at the origin $n=0$. If we take note of the fact that the there is a pair of ions with negative charges at $n=1$ and $-1$, a pair of ions with position charges at $n=2$ and $-2$, and onward, we can arrive at the expression
$$U = \frac{-2e^2}{4a\pi\epsilon} \sum_{n=1}^{\infty} \frac{1}{n}(-1)^{n-1} \, .$$
The 2 in front of the $e^2$ accounts for the fact that the charges are in pairs. We can then recognize that the summation is a Taylor series approximation for $\ln(2)$. Hence, the final answer is potential energy
$$\frac{-\ln(2) e^2}{2a\pi\epsilon}$$
possessed by this ion. All the other ions should possess the same energy.
Method 2
Do everything done in step 1, but multiply the expression by $N$ (total number of particles in the lattice) to calculate the total potential energy of the entire system setting each point as the origin:
$$U=\frac{N}{2}\frac{-2e^2}{4a\pi\epsilon} \sum_{n=1}^{\infty} \frac{1}{n}(-1)^{n-1} \, .$$
The total number of particles is divided by two to avoid double counting of the energy between particles (between particle 0 and 1 and between 1 and 0, say). The total energy of the system is then divided by $N$ again to find the energy per ion. The final answer is
$$\frac{-\ln(2)*e^2}{4a\pi\epsilon} \, .$$
For what it's worth, the two links below calculate the summation as a Madelung constant and arrives at the answer shown in method 1. Perhaps there is a distinction between energy per ion and energy of an ion that I am missing?

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*http://web.unbc.ca/~hussein/Phys_206_Winter_2004/Phys_206_Ch11_HW.pdf


*http://homerreid.dyndns.org/teaching/18.330/Notes/Invitation.pdf
Method 2 is based on a similar method shown in Purcell's E&M text.
 A: For simplicity, think about the case with just two charges, where the potential energy is 
$$U_{\text{system}}(\mathbf{r}_1, \mathbf{r}_2) = \frac{k q^2}{|\mathbf{r}_1 - \mathbf{r}_2|}.$$
This potential energy is a property of the entire system, not either charge alone. This is the energy you could harvest if you took the entire system apart.
However, you could also describe this potential energy as a property of the first charge,
$$U_1(\mathbf{r}_1) = \frac{k q^2}{|\mathbf{r}_1 - \mathbf{r}_2|}.$$
This would be a natural thing to do if the second charge were very heavy, or otherwise unable to move. For example, we think precisely in this way when we write the gravitational potential energy of a mass on the Earth as $mgh$. It really comes from the gravitational potential energy of the system of the Earth and mass, but it's useful to think of it as a property of the mass because the Earth is essentially fixed. 
Both methods you've shown are perfectly correct, but they compute different things. The first computes the potential energy of a given charge, treating all the other charges as a fixed background. It is equal to the amount of energy it would take to remove that charge alone to infinity. The second computes the potential energy of the entire system divided by the number of charges. It is equal to the amount of energy it would take to move all charges to infinity, divided by the number of charges. Of course we expect it to be harder to move the first charge away, so the answer in method $1$ is larger. With a little thought, you can see that it must be precisely twice as large.

Perhaps there is a distinction between energy per ion and energy of an ion that I am missing?

Yes. When you speak of the energy per ion, you usually mean the energy of the system divided by the number of ions. When you speak of the energy of an ion, you usually mean the energy of that ion alone, regarding the rest of the system as fixed. Unfortunately, the usage isn't entirely standard.
