Radius of Convergence Is the radius of convergence of 
$$\frac{n(x+3)^n}{4^n}$$
equals 4?
I got $|x+3|\lt 4$ as the final result. How do you know, what is the radius from here? 

The radius of convergence of 
$$
\sum_{n=0}^\infty a_n(x-a)^n\tag{1}
$$
is the number $R$ such that (1) converges for $\lvert x-a\rvert<R$ and diverges for $\lvert x-a\rvert>R$. In our case, we can apply the Ratio Test to the power series
$$
\sum_{n=0}^\infty \frac{n}{4^n}(x+3)^n\tag{2}
$$ 
Doing so gives
$$
\lim_{n\to\infty}\left\lvert\frac{\frac{n+1}{4^{n+1}}(x+3)^{n+1}}{\frac{n}{4^n}(x+3)^n}\right\rvert
= \lim_{n\to\infty} \frac{n+1}{n}\frac{4^n}{4^{n+1}}\left\lvert\frac{(x+3)^{n+1}}{(x+3)^n}\right\rvert 
= \frac{1}{4}\lim_{n\to\infty}\frac{n+1}{n}\left\lvert x+3\right\rvert
=\frac{1}{4}\left\lvert x+3\right\rvert
$$
It follows that (2) converges when
$$
\frac{1}{4}\left\lvert x+3\right\rvert<1
$$
and diverges when
$$
\frac{1}{4}\left\lvert x+3\right\rvert>1
$$
Equivalently, (2) converges when 
$$
\left\lvert x+3\right\rvert<4
$$
and diverges when
$$
\left\lvert x+3\right\rvert>4
$$
Thus the radius of convergence of (2) is $R=4$.
