Does the potential of a charged ring diverge on the ring? I know that the density and potential (in spherycals) of a charged ring is, respectively,:
$$
\rho(\textbf{r})
=
\frac{\lambda}{a}
\delta(r-a)\delta(\theta-\tfrac{\pi}{2})
$$
$$
\varphi(\textbf{r})=
\frac{2\pi a \lambda}{r_>}
\left[
 1+    \sum_{n=1}^\infty (-1)^n \frac{(2n-1)!!}{(2n)!!}\left(\frac{r_<}{r_>}\right)^{2n}P_{2n}(\cos\theta)
\right]
$$
Where $P_{2n}$ is the $2n$-th Legendre Polynomial, and $r_<=\min\{a,r\},r_>=\max\{a,r\}$. If I evaluate $\mathbf r$ in the ring ($r=a,\theta=\tfrac{\pi}{2}$):
$$
\varphi(\mathbf r)\,\propto\,
\left[
1+    \sum_{n=1}^\infty (-1)^n \frac{(2n-1)!!}{(2n)!!}
\right]
\to\infty
$$
So this is a problem (I suppose).
 A: This is actually a fun question, I learnt something new about double factorials while trying to answer it! 
I don't see why that term diverges. Using the identities on Wikipedia for the "double factorial", we have that for even integers $k$,
$$\int_0^{\pi/2} \sin^{k}(x)\text{d}x = \frac{(k-1)!!}{(k)!!}\frac{\pi}{2}.$$
We can use this to calculate the sum term you have explicitly.
$$\sum_{n=1}^\infty (-1)^n \frac{(2n-1)!!}{(2n)!!} = \frac{2}{\pi}\sum_{n=1}^\infty (-1)^n \int_0^{\pi/2} \sin^{2n}(x)\text{d}x = \frac{2}{\pi}\int_0^{\pi/2} \text{d}x \sum_{n=1}^\infty (-1)^n \sin^{2n}(x).$$
Where in the last step I've interchanged the sum and the integral. This particular sum is quite easy to do, and I'll leave it as an exercise to show that 
$$\sum_{n=1}^\infty (-1)^n \sin^{2n}(x) = -\frac{\sin^2(x)}{1+\sin^2(x)}.$$
We can now perform the integral and show that $$-\frac{2}{\pi}\int_0^{\pi/2} \frac{\sin^2(x)}{1+\sin^2(x)} \text{d}x = \frac{-2 + \sqrt{2}}{2}.$$
Thus, $$\sum_{n=1}^\infty (-1)^n \frac{(2n-1)!!}{(2n)!!} = \frac{-2 + \sqrt{2}}{2} < \infty,$$
which should solve your problem. 
