2. 1) Let $\{ a_n\}$ be be complex sequence. Say $w\not = 0$ is such that $\{ |a_nw^n| : n\geq 0\}$ is bounded. Let $A$ be an upper bound, so, $|a_nw^n| < A$ for all $n\geq 0$. Let $|z| < |w|$. Note, $|a_nz^n| \leq |a_nw^n| \cdot |\tfrac{z}{w} |^n < A \left| \tfrac{z}{w} \right|^n$. But $\left|\tfrac{z}{w}\right|^n$ is a geometric series with $|z/w|<1$ so it converges. The radius of convergence is define to be $R$ ( $R=0$ or $R=\infty$ are included) so that if $|z| then we have convergence and if $|z|>R$ then we have divergence.
2)If $\lim |a_n|^{1/n} = R$ then it means $\lim |a_n|^{1/n} |z| = R|z|$. Thus, if $|z| < 1/R$ then $R|z| < 1$. Pick $\epsilon > 0$ to that $R |z| + \epsilon < 1$. Then for sufficiently large $n$ we have that $||a_n|^{1/n}|z| - R|z|| < \epsilon \implies R|z| - \epsilon< |a_n|^{1/n} |z| < R|z|+\epsilon$. Thus, $|a_nz^n| < (R|z|+\epsilon)^n$ but $0 so the geometric series converges. By direct comparison test it follows that $|a_nz^n|$ converges for $|z|<1/R$. Similarly if $|z|>1/R$ then $\lim |a_n|^{1/n}|z| = R|z| > 1$. Pick $\epsilon > 0$ so that $R|z| - \epsilon > 1$ and so as in above $|a_nz^n| > (R|z| - \epsilon)^n$ and this is a divergent geometric series.
3)For (a) consider $\sum_{n=0}^{\infty} \frac{z^n}{n^2}$ using #2 the radius of convergence is $1$. If $z$ is any point on the circle then $\left| \frac{z^n}{n^2} \right| \leq \frac{1}{n^2}$ so it converges too. For (b) consider $\sum_{n=0}^{\infty} z^n$, it does not converge on the circle because if $|z| = 1$ then $\lim |z^n| = 1\not = 0$, so by the divergence test it cannot converge. For (c) consider $\sum_{n=0}^{\infty} \frac{(-1)^nz^n}{n}$. At $z=1$ it converges but not at $z=-1$.