The given power series is: $$\sum^\infty_{n=1}\frac{(x+1)^n}{n3^n}$$ In order determine the endpoints of the interval we first need to take the limit of the sum using the ratio test: $$\lim_{n\to\infty} \left|\frac{(x+1)^{n+1}}{(n+1)3^{n+1}} \bullet \frac{n3^n}{(x+1)^n} \right|$$ After simple elimination and factoring we get: $$\left| \frac{x+1}{3}\right|\lim_{n\to\infty} \frac{n}{n+1}= \left| \frac{x+1}{3}\right|$$ Then determining the endpoint is easy, because the convergence depends on the value of $x$ which we can deduce from the inequality: $$\left| \frac{x+1}{3}\right| <1$$ $$\left|{x+1}\right|<3$$ $$-2<x<2$$ Solving for $x=-2$ within the given: $$\sum^\infty_{n=1}\frac{(-2+1)^n}{n3^n}=\sum^\infty_{n=1}\frac{(-1)^n}{n3^n}$$ Now by the alternating series test $b_n=\frac{1}{n3^n}$, and two conditions must be satisfied: $$1) \lim_{n \to \infty}\frac{1}{n3^n}=0 $$ $$2)\space b_{n+1}< b_n$$ and both are true, so the right end point converges.
Now for the left end point:$x=2$
$$\sum^\infty_{n=1}\frac{(2+1)^n}{n3^n}=\sum^\infty_{n=1}\frac{3^n}{n3^n}=\sum^\infty_{n=1}\frac{1}{n}$$
Using the integral test we find the endpoint to be divergent:
$$\int_1^\infty \frac{1}{x}dx=\ln(x)$$
$$\ln(\infty) -\ln(1)= \infty$$
So the interval of convergence is:
$$[-2,2)$$
I post this question only looking to have my work checked over as my professor has been difficult to communicate with over email due to COVID-19.
$|x+1|<3$ is same as $-3<x+1<3$ or $-4 <x<2$. The interval is $[-4,2)$ and your arguments are all basically correct. Just change the left end point of the interval.