Laplace equation derivation with possible omission of an addition In this MIT video on finding the Laplace equation on a circle of $r=1$ with a point source of temperature at $r=1,\,\theta =0,$ the speaker gets to a "punch line" of the whole presentation when he says that the value of the equation at $r=0$ along the $x$ axis is $\frac{1}{2\pi}:$


Well, of course, it's the average value: right at the center, the temperature is going to be the average on the boundary.

However, there seems to be a plus sign missing between $\frac{1}{2\pi}$ and the rest of the expression at
$$\frac{1}{2\pi}\frac{1-r^2}{1+r^2-2r\cos\theta}$$
unless I'm not understanding how the geometric series is applied.
If the plus is indeed missing, the value at $(r,\theta)=(0,0) $ should be $\frac{1}{2\pi}+1!$

How can this be reconciled?

 A: Note that for $0\le r<1$ we have 
$$\begin{align}
\frac1{2\pi}+\frac1\pi\sum_{n=1}^\infty r^n\cos(n\theta)&=\frac1{2\pi}+\frac1\pi\text{Re}\left(\sum_{n=1}^\infty (re^{i\theta})^n\right)\\\\
&=\frac1{2\pi}+\frac1\pi\text{Re}\left(\frac{re^{i\theta}}{1-re^{i\theta}}\right)\\\\
&=\frac1{2\pi}+\frac1\pi \text{Re}\left(\frac{re^{i\theta}(1-re^{-i\theta})}{1+r^2-2r\cos(\theta)}\right)\\\\
&=\frac1{2\pi}+\frac1\pi \frac{r\cos(\theta)-r^2}{1+r^2-2r\cos(\theta)}\\\\
&=\frac1{2\pi}-\frac1{2\pi}\frac{2r^2-2r\cos(\theta)}{1+r^2-2r\cos(\theta)} \\\\
&=\frac1{2\pi}-\frac1{2\pi}\frac{r^2 -1+r^2-2r\cos(\theta)+1}{1+r^2-2r\cos(\theta)} \\\\
&=\frac1{2\pi}\frac{1-r^2}{1+r^2-2r\cos(\theta)}\tag1
\end{align}$$
as was obtained in the development that was referenced in the OP.
Obviously, the expression on the right-hand side of $(1)$ has the same limit (i.e., $1/2\pi$) as $r\to 0$ as the limit of the original expression $\lim_{r\to 0}\left(\frac1{2\pi}+\frac1\pi\sum_{n=1}^\infty r^n\cos(n\theta)\right)=\frac1{2\pi}$
