Solution to Laplace's equation on the disk I am reading through this document http://web.math.ucsb.edu/~grigoryan/124B/lecs/lec10.pdf, which arrives at the following series solution to Laplace's equation on the disk:
$$u(r, \theta)=\frac{A_{0}}{2}+\sum_{n=1}^{\infty} r^{n}\left(A_{n} \cos n \theta+B_{n} \sin n \theta\right).$$


*

*Why do we sum up all the eigenfunctions? How do we know this gives a solution? (I know very little about diff. equations, and I'm just looking for a brief explanation here.)

*What is the nature of the convergence of this series? Can we pick $A_n$ and $B_n$ that decay rapidly enough so that  it converge uniformly on every compact subset? 
 A: The other answer notwithstanding, this is not "just a formal solution", it is the actual solution to the Dirichlet problem in the unit disk.
Let's say $f\in C(\Bbb T)$ just to keep things simple (that is, $f$ is continuous and has period $2\pi$). We want a function $u(r,t)$ which is continuous in the closed unit disk $\overline D$, is harmonic in $D$, and which equals $f$ on the boundary.
The notation is simpler if we use the "complex" form of Fourier series. Let $$c_n=\hat f(n)\quad(n\in\Bbb Z).$$The $c_n$  are certainly bounded, so the series $$u(r,t)=\sum_{n=-\infty}^\infty c_nr^{|n|}e^{int}$$converges uniformly on compact subsets of the unit disk.
Since $z^n$ and $\left(\overline z\right)^n$ are analytic it follows that the partial sums are harmonic. Since a uniform limit of harmonic functions is harmonic this shows that $u$ is harmonic in the (open) disk.
To answer a question you ask in a comment, the Fourier series of $f$ need not converge. But that doesn't matter. Look up "Abel summability" somewhere; it's very well  known that the Fourier series of $f$ is uniformmly Abel summable to $f$; this means precisely that $u(r,t)\to f(t)$ uniformly as $r\to 1$.
About convergence to $f$ at the boundary: Say $P_r$ is the Poisson kernel. It's  well known and not that hard to prove that the convolution $P_r*f$ tends to $f$ uniformly. It's also well known that $$P_r(t)=\sum_{n=-\infty}^\infty r^{|n|}e^{int}$$(you can calculate that sum explicitly, since it's just the sum of two geometric series), which makes it clear that $$P_r*f(t)=u(r,t).$$
Come to think of it, since the title asks about harmonic functions without mentioning boundary values, it may be appropriate to point out that in fact any function $u(r,t)$ harmonic in the open unit disk is given by a series as above for some choice of coefficients $c_n$.
Outline of proof: If $u$ is a real-valued harmonic function in the open disk we know there exists a real-valued harmonic function $v$ such that $u+iv$ is analytic. This shows that if $u$ is any harmonic function in the disk there exist analytic functions $f$ and $g$ such that $$u=f+\overline g.$$Now $f$ and $g$ are both given by power series...
A: This is just a formal solution. If you just differentiate termwise ignoring any convergence issues this should come out as an eigenfunction. The tricky part is now to find which coefficients to pick for $A_n$ and $B_n$ to satisfy the boundary conditions and then prove some kind of convergence.
