Elementary convergence of power series 
Prove that the power series $\sum_{j=0}^{\infty}z^j$ converges at no
  point on its circle of convergences $|z|=1$. Also, prove that the
  power series $\sum_{j=0}^{\infty}\frac{z^j}{j^2}$ converges at every
  point on its circle of convergences $|z|=1$.

This question is confusing to me. I know that a geometric series is divergent when $|z|>1$ and convergent when $|z|<1$, but what happens at the boundaries of the circle $|z|=1$? Also, the second part of the question is confusing too, I know that if I just use the comparison test that it will converge, but I don't think that is correct? What will happen if $\sum_{j=0}^{\infty}\frac{z^j}{j^2}$, where $|z|>1$ or $|z|<1$? 
 A: For the first, since the $n$th term is $z^n$ and $|z^n| = |z|^n = 1$, we see that the terms do not converge to zero, hence the series cannot converge.
For the second, the $n$th term is $\frac{z^n}{n^2}$ and $|\frac{z^n}{n^2}| = \frac{1}{n^2}$ which is a convergent series, hence it converges absolutely for all $z$ such that $|z|=1$.
Since $\frac{1}{R} = \limsup_n \sqrt[n]{\frac{1}{n^2}} = \limsup_n e^{-\frac{2}{n} \ln n} = 1$, we see that the second series converges for $|z| <1$, and diverges for $|z|>1$.
Alternatively, you could use the ratio test: Since  $\rho =\lim_n |\frac{a_{n+1}}{a_n}| = \lim_n |\frac{\frac{z^{n+1}}{(n+1)^2}}{\frac{z^{n}}{n^2}}| = |z|$, we see that the series converges for $|z| <1$ and diverges for $|z|>1$.
A: For the first use that for $|z|=1$ $z^k$ doesn't converge to zero and hence the series can't be convergent,  for the second use comparism to $\sum_{k=1}^\infty \frac{1}{k^2}$.
When $|z|>1$ the sequence doesn't converge to zero and so the series can't be convergent, for $|z|<1$ the comparism gives us the absolut convergence.
A: Hint: the second series is absolutely convergent for $|z|=1$ since $\sum_n\frac{1}{n^2}$ is convergent.
For the first series: it diverges because $z^n=e^{in\theta},$ don't converge to $0$ for all $\theta$.
