$\require{cancel}$
Find the interval of convergence for:
$$\sum_{i=1}^\infty \frac{(x+2)^n}{(n^{1/2})(3^n)}$$
Progress
I used the root test,
$$L=\lim\limits_{n \to \infty} \sqrt[n]{\frac{(x+2)^n}{(n^{1/2})(3^n)}}=\frac{x+2}{\cancel{(n^{1/2n})}(3)}=\frac{x+2}{3}$$
Therefore, to discover the interval of convergence I did this:
$$\lvert\frac{x+2}{3}\rvert<1$$
Thus, getting this:
$$-1<\frac{x+2}{3}<1\implies-5<x<1$$
The problem here is, I don't remember how to check if the endpoints converge or not. I tried substituting each endpoint into the original series.
- For $x=-5$, I discovered that it was an alternating series, where $\sum_{i=1}^\infty |a_n|$ diverges, but $\sum_{i=1}^\infty a_n$ converges [conditionally], assuming $a_n=\frac{(x+2)^n}{(n^{1/2})(3^n)}$.
- For $x=1$, I found that $\sum_{i=1}^\infty a_n$ diverges.
How can I determine if the endpoint converges or not? Must it converge (at that $x$ value) absolutely, or conditionally?
Best Answer
As suggested, I'm posting my comment as an answer so we can clear this one off the plate.
You're done, the interval is $[−5,1)$. It doesn't have to converge absolutely at the endpoints and it typically doesn't, it often converges only conditionally at the endpoints, if at all but that doesn't change the interval of convergence. Conditionally convergent endpoints are still part of the interval of convergence.