Find radius of convergence of $\sum_{n=0}^{\infty} (2n+1)z^n$ Let $$\sum_{n=0}^{\infty} (2n+1)z^n$$ be a power series with $z \in \mathbb{C}$ 
Find the radius of convergence and show that the series is divergent for all $z \in \mathbb{C}$ on the boundary

So by using the ratio test I have found the radius of convergence to be equal to 1. For $z=-1$ and $z=1$ it's fairly straightforward to see the series diverges. Also I can see that for z with modulus 1:
$$\sum_{n=0}^{\infty} \mid (2n+1)z^n \mid = \sum_{n=0}^{\infty} (2n+1) \mid z \mid^n =\sum_{n=0}^{\infty} (2n+1) = \infty$$
So we have "absolute divergence". However as far as I know this does not imply regular divergence... So my question is how do I check for all other complex numbers on the boundary apart from 1 and -1?
 A: For $|z|=1$ we have that the sequence $((2n+1)z^n)$ does not converge to $0$, since
$|(2n+1)z^n|=2n+1 \to \infty$ as $n \to \infty.$ Hence $ \sum_{n=0}^{\infty} (2n+1)z^n$ is divergent for $|z|=1.$
A: When $|z|=1$ you can write $z = e^{i \theta} = \cos \theta + i \sin \theta$ and the  series become
$$
\sum_{i=0}^{\infty}(2n+1)(\cos (n \theta) + i \sin (n \theta))= \sum_{i=0}^{\infty}(2n+1)\cos (n \theta)  + i \sum_{i=0}^{\infty}(2n+1)\sin (n \theta),
$$
which is divergent (for any given $\theta$, at least one of real or imaginary parts are divergent series).
A: In this case, I would use the fact that $ \limsup_n |(2n+1)z^n| = \infty \neq 0$ which is necessary for the series to converge (why?). 
Try to adapt this argument to the convergence on the boundary of the power series $\sum_{n \ge 0 } z^n$ as an exercise if you wish.
Let us now look at a slightly less trivial example, say power series with a general term $\frac{z^n}{2n+1}$. The convergence radius is the same, the convergence on the boundary is not. Moreover, the series diverges absolutely for $|z|=1$. 
One can pass to polar coordinates, if $|z|=1$ then $z= \exp(i \theta)$ for some real parameter $\theta \in [0, 2 \pi [$ which yields $$\sum_n \frac{\exp(ni \theta)}{2n+1}.$$
Now for $z \neq 1$ (i.e. $\theta \neq 0 $) we can apply Abel's test (explain why we can do so, noting that in this case, $\sum_{N\ge n \ge 0}\exp(ni\theta)$ is bounded independently of $N$. See here for example.).

P.S. One famous but more advanced theorem in the direction of your question:
https://en.wikipedia.org/wiki/Fatou%27s_theorem
