Determine the radius of convergence 
of the power series:
![\[ \sum_{n=0}^{\infty} \frac{(z+3)^n}{(n+1)2^n}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ccb4482d262f87c61a40e723065d7538_l3.png "Rendered by QuickLaTeX.com")
Test for convergence at the boundary points if is finite.
Let 
. Then
![\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{(z+3)^{n+1}}{(n+2)2^{n+1}} \right) \left( \frac{(n+1)2^n}{(z+3)^n} \right) \\[9pt] &= \lim_{n \to \infty} \frac{(z+3)(n+1)}{2(n+2)}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-8bb8baefaff0085204fb7aff08185cef_l3.png "Rendered by QuickLaTeX.com")
This implies the series converges for 
which implies 
. Furthermore, the series converges on the boundary except the point 
. Hence, the series converges for all 
such that 
and 
.
I think there is a typo in the conclusion: the series converges for all $z$ such that $|z+3| \leq 2$, not 3