This problem is from complex analysis.
Set $$f(z)=\sum_{n=0}^{\infty}a_nz^n$$ with convergence radius of 1, and $$\lim_{n \to \infty}a_n=0$$ Prove that if $z_0 \in \partial B(0,1)$ is not a singular point of f(z), then $$\sum_{n=0}^{\infty}a_n z_0^n$$ converges.
It's easy to see that if $$\lim_{n \to \infty}na_n=0$$, we can quickly solve it because of Tauber's theorem. But now the condition $$\lim_{n \to \infty}a_n=0$$ seems too weak, is it really sufficient enough for the problem?
Assume that $z_0=1$ and $f(1)=0$. Let $s_n=\sum_{k=0}^n a_k$. We prove that $ s_n\to 0$ as $n\to \infty$.
For $r<1$ we have $$ s_n=\frac{1}{2\pi i}\int_{|z|=r} \frac{f(z)}{(1-z)z^{n+1}}dz=\frac{1}{2\pi r^n}\int_{-\pi}^\pi \frac{f(re^{i\theta })e^{-in\theta }}{1-re^{i\theta }}d\theta .$$
Since $f(z)$ is analytic at $z=1$ and $f(0)=0$, we see that $\frac{f(z)}{1-z}$ is also analytic at $z=1$. Therefore for arbitrary $\varepsilon >0$ there exists $\delta >0$ such that $$ \left| \frac{1}{2\pi r^n}\int_{-\delta }^\delta \frac{f(re^{i\theta })e^{-in\theta }}{1-re^{i\theta }}d\theta\right|<\frac{\varepsilon }{3\,r^n}\quad (0<r\le 1) .$$
Now consider a (not analytic) function $\phi(z;\delta )\in C^2 (|z|\le 1)$ with \begin{align} &\phi(z;\delta )=\frac{1}{1-z}\quad \text{for}\; z=re^{i\theta },\, r\le 1,\, \delta \le |\theta |\le \pi,\\ &|\phi|,\, \left|\frac{\partial \phi}{\partial \theta }\right|, \, \left|\frac{\partial^2 \phi}{\partial \theta^2 }\right|\le K(\delta )\; \text{for}\;\text{some}\;\text{constant}\; K(\delta ). \end{align}
Then we have \begin{align} s_n=\frac{1}{2\pi r^n}&\left\{\int_{-\delta }^\delta \frac{f(re^{i\theta })e^{-in\theta }}{1-re^{i\theta }}d\theta +\int_{-\pi}^\pi f(re^{i\theta })\phi(re^{i\theta};\delta )e^{-in\theta }d\theta \right. \\ &\left. -\int_{-\delta }^\delta f(re^{i\theta })\phi(re^{i\theta} ;\delta )e^{-in\theta}d\theta\right\} . \end{align}
Integrating by parts twice we have (to change the order of integral and summation, note that the convergence is uniform for $r<1$) \begin{align} &\int_{-\pi}^\pi f(re^{i\theta })\phi(re^{i\theta};\delta )e^{-in\theta }d\theta\\ &=\int_{-\pi}^\pi \left(\sum_{k=0}^\infty a_kr^ke^{ik\theta } \right)\phi(re^{i\theta};\delta )e^{-in\theta }d\theta\\ &=\sum_{k=0}^\infty a_kr^k\int_{-\pi}^\pi \phi(re^{i\theta};\delta )e^{-i(n-k)\theta }d\theta\\ &=a_nr^n\int_{-\pi}^\pi \phi(re^{i\theta };\delta )d\theta +\sum_{k=0,k\ne n}^\infty \int_{-\pi}^\pi \phi(re^{i\theta };\delta )e^{-i(n-k)\theta }d\theta \\ &=a_nr^n\int_{-\pi}^\pi \phi(re^{i\theta };\delta )d\theta-\sum_{k=0,k\ne n}\frac{a_kr^k}{(n-k)^2}\int_{-\pi}^\pi \frac{\partial^2 \phi}{\partial\theta ^2}\cdot e^{-i(n-k)\theta }d\theta . \end{align} Here by the properties of $\phi$ we see that \begin{align} &\left|\int_{-\pi}^\pi \phi(re^{i\theta };\delta )d\theta\right|\le 2\pi K(\delta ),\\ &\left|\int_{-\pi}^\pi \frac{\partial^2 \phi}{\partial\theta ^2}\cdot e^{-i(n-k)\theta }d\theta\right| \le 2\pi K(\delta ). \end{align}
Thus we have \begin{align} &\left|\int_{-\pi}^\pi f(re^{i\theta })\phi(re^{i\theta};\delta )e^{-in\theta }d\theta\right|\\ &\le 2\pi K(\delta )\left(|a_n|+\sum_{k=0,k\ne n}^\infty \frac{|a_k|}{(n-k)^2} \right)\\ &=2\pi K(\delta )\left(|a_n|+\sum_{k\le n/2} \frac{|a_k|}{(n-k)^2} + \sum_{n/2<k,k\ne n}^\infty \frac{|a_k|}{(n-k)^2} \right)\\ &\le 2\pi K(\delta )\left(|a_n|+\frac{2}{n}\max_{k\le n/2} |a_k| +2 \max_{n/2<k} |a_k|\cdot \sum_{m=1}^\infty \frac{1}{m^2} \right)\\ &< \frac{2\pi \varepsilon }{3} \end{align} for sufficiently large $n$, since $|a_n|\to 0$ as $n\to \infty$.
Similarly by integration by parts we have \begin{align} \left|\int_{-\delta }^\delta f(re^{i\theta })\phi(re^{i\theta} ;\delta )e^{-in\theta}d\theta\right|&= \left| \left[f\phi\frac{e^{-in\theta }}{-in}\right]_{-\delta }^\delta +\frac{1}{in}\int_{-\delta }^\delta \frac{\partial(f\phi)}{\partial\theta }e^{-in\theta }d\theta \right|\\ &< \frac{2\pi\varepsilon }{3} \end{align} for sufficiently large $n$.
Thus we have $$ |s_n|< \frac{\varepsilon }{r^n} \quad (0<r<1)$$ for sufficiently large $n$. Letting $r\to 1$ we have $|s_n|\le \varepsilon $, which ensures $s_n\to 0\; (n\to \infty)$.
The proof is complete.