Show that a necessary condition for the existence of a solution to the Neumann problem \begin{cases} \frac{\partial^2 u}{\partial r^2}+ \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} = 0 ,\hspace{0.4cm} 0<r<a, \hspace{0.2cm} -\pi < \theta \leq \pi\\ \frac{\partial u}{\partial r}(a , \theta) = f(\theta) ,\hspace{0.4cm} -\pi < \theta \leq \pi \end{cases} is that $$\int_{-\pi}^{\pi}{f(\phi)d \phi} =0$$ that is, the mean value of the normal derivative at the boundary is zero.
My attempt:
I'm trying to test for this problem. If I'm not wrong, then the general solution to this problem is $$u(r,\theta) = \frac{A_0}{2} + \sum_{n=1}^{\infty}{r^{n}\left(A_{n}\cos(n\theta) + B_{n}\sin(n\theta) \right)} \hspace{0.8cm} (*) $$ Then, assuming uniform convergence, I get that $$\frac{\partial u}{\partial r}(r ,\theta) = \sum_{n=1}^{\infty}{nr^{n-1}\left(A_{n}\cos(n\theta) + B_{n}\sin(n\theta) \right)}$$ The solution $(*)$ satisfies the problem condition if $$f(\theta) = \frac{\partial u}{\partial r}(a ,\theta) = \sum_{n=1}^{\infty}{na^{n-1}\left(A_{n}\cos(n\theta) + B_{n}\sin(n\theta) \right)}$$ Therefore integrating both sides and assuming that I can interchange the series with the integration, we have that \begin{align} \int_{-\pi}^{\pi}f(\phi)d\phi &= \int_{-\pi}^{\pi}\sum_{n=1}^{\infty}{na^{n-1}\left(A_{n}\cos(n\theta) + B_{n}\sin(n\theta) \right)}d\phi \\ &=\sum_{n=1}^{\infty}na^{n-1}\int_{-\pi}^{\pi}{\left(A_{n}\cos(n\theta) + B_{n}\sin(n\theta) \right)}d\phi\\ &= \sum_{n=1}^{\infty}na^{n-1}{\left(A_{n}\int_{-\pi}^{\pi}\cos(n\theta)d\phi + B_{n}\int_{-\pi}^{\pi}\sin(n\theta) d\phi \right)}\\ &=0 \end{align} I can notice that $\sum_{n=1}^{\infty}{na^{n-1}} = \frac{1}{(a-1)^2} \hspace{0.3cm}$ if $|a|<1$
I'm not very satisfied with this reasoning and I feel that I have made a serious mistake. So I decided to post this statement and get to know your most expert answers. I appreciate some help.
$\textbf{Hint:}$ An alternative way to show this is to write the Laplace equation as $\Delta u=\nabla\cdot\nabla u=0$ and apply the divergence theorem.
EDIT: Define $D=\{(r,\theta)\,|\,0<r<a,\,-\pi<\theta<\pi\}$ and integrate the Laplace equation over $D$. This gives $$0=\iint_D \Delta u\,dA=\iint_D\nabla\cdot\nabla u\,dA=\int_{\partial D}\nabla u \cdot\mathbf{\hat{r}}\,ds$$ where the last equality holds by the divergence theorem in two dimensions. $dA$ is the area element, $ds$ is the line element along the boundary, and $\mathbf{\hat{r}}$ is the unit normal of the circle. But $\nabla u \cdot\mathbf{\hat{r}}=\frac{\partial u}{\partial r}$ is precisely the normal derivative. So by the boundary condition, we have $$0=\int_{\partial D}\frac{\partial u}{\partial r}\,ds=\int_{\partial D}f(\theta)\,ds.$$