Codex @codex  0

Convergence at $z_0$ implies that the terms $a_n{z}_0^n$ form a bounded sequence: $|a_n{z}_0^n|\le M$ for some $M$. If $|z|<|z_0|$ and $r=|z/z_0|<1$, thenComparison with the convergent geometric series $\sum M{r}^n$ proves absolute convergence. The case $z_0=0$ is vacuous.

The radius of convergence is the number $R\in [0,\infty ]$ such that a power series converges absolutely for $|z|<R$ and diverges for $|z|>R$, with boundary behavior considered separately.

TakeThen $\sum v^n/n$ is the convergent alternating harmonic series, whereas $\sum w^n/n$ is the divergent harmonic series. Thus $\sum z^n/n$ converges at a point of modulus one and diverges at another point of modulus one. It follows that

The alternating series test says that if $a_n\ge 0$, $a_{n+1}\le a_n$, and $a_n\to 0$, then ${\sum }_{n\ge 1}(-1{)}^{n+1}{a}_n$ converges.

To prove it, let $S_N$ be the partial sums. The even sums satisfyso $(S_{2m})$ is increasing. The odd sums satisfyso $(S_{2m+1})$ is decreasing. Also $S_{2m}\le S_{2m+1}$, so both are bounded and converge. Their difference is $a_{2m+1}\to 0$, hence their limits agree and the whole sequence $(S_N)$ converges.

Taking $a_n=1/n$ proves convergence of the alternating harmonic series. Its even partial sums lie below its limit $S$, while its odd partial sums lie above it. Sinceandwe obtain