TITLE: Series Radius of Convergence (Absolute)
QUESTION [0 upvotes]: $$
\sum_{j=1}^{\infty}(-1)^{n+1}\left(\frac{n+1}{n^2}\right)x^n.
$$
Find the real numbers $x$ for which the series $(i)$ converges and $(ii)$ absolutely converges.
I did the ratio test and found radius of convergence to be $1$. I checked endpoints and I ended up with $x \in (-1,1]$. I believe this is correct but I might be wrong. I know this is the set of values of $x$ for which the series absolutely converges so is the answer to $(ii)$.
So for $(i)$ what would I do. I know if a series is absolutely convergent then it is convergent, so would the answer be the same? My other thought was to when doing $\frac{a_{n+1}}{a_n}$ to drop the absolute values. Either way I end up with the same answer so would this be the difference in working out I would have to show ?

REPLY [3 votes]: For $x = 1$, the series converges by the alternating series test, but does not converge absolutely by comparison with $\sum_n 1/n$.  For $x = -1$, it neither converges nor converges absolutely.  So $(-1,1]$ is the answer to (i) and $(-1,1)$ is the answer to (ii).
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