A theorem on Taylor series in complex domain is as follow:
Suppose $f(z)$ has Taylor series at $a$ with convergence radius of $R$. Then $f(z)$ has at least one singular point on $|z-a|=R$.
But I find an example as follow: $$ f(z)=\sum_{n=2}^{\infty}\frac{z^n}{n(n-1)} $$ that has $R=1$. Moreover $f(z)$ is uniform convergent for all $|z|\leqslant1$ since $$ |f(z)|\leqslant \sum_{n=2}^{\infty}\frac{1}{n(n-1)}<1 $$ Since $z^n$ is analytic on $|z|\leqslant1$, by Weierstrass theorem, $f(z)$ is analytic on $|z|\leqslant1$ too. Moreover it converges on entire $|z|=1$, and diverges for $|z|>1$. By taking derivative, we have $$ f''(z)=\sum_{n=0}^{\infty}z^n=\frac1{1-z} $$ By integrating twice we can get $$ f(z)=(1-z)\log(1-z)+z. $$ but only on $|z|<1$. Can anyone explain this? Is the above theorem wrong?
The Weierstrass theorem you are referring to only refers to open sets: if $f_n$ is a sequence of analytic functions on an open set which converges uniformly to a function $f$, then $f$ is analytic. So to conclude that the limit is analytic on $|z|\leq 1$ in this case (which really means $f$ is analytic in a neighborhood of $|z|\leq 1$, since "analytic" only makes sense on open sets), you would need to have uniform convergence not just on $|z|\leq 1$ but on an open neighborhood of it.
In the statement of the theorem you are asking about, to say $f$ is "singular" at a point $p$ means that $f$ cannot be extended to an analytic function in a neighborhood of $p$. In this case, you can see from your formula that $f$ is singular at $p=1$, even though $f$ can be defined continuously on the entire set where $|z|\leq 1$.