TITLE: Exercise asks to show that $\sum_{n=1}^{\infty}\frac{n+1}{n}\frac{1}{3^n}x^n$ converges for $x\in(-1,1)$. I show $x\in(-3,3)$. Is something wrong?
QUESTION [5 upvotes]: Consider the power series $\sum_{n=1}^{\infty} \frac{n+1}{n}\frac{1}{3^n}x^n$.
1. Show that the series converges for all $x\in (-1,1)$.
If we let $a_n = \frac{n+1}{n}\frac{1}{3^n}x^n$ then using the ratio test we get
\begin{equation*}
\begin{split}
\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| &= \lim_{n\to\infty} \left|\frac{\frac{n+2}{n+1}\frac{1}{3^{n+1}}x^{n+1}}{\frac{n+1}{n}\frac{1}{3^n}x^n}\right| \\
&= \lim_{n\to\infty} \left|\frac{n+2}{n+1}\frac{1}{3^{n+1}}x^{n+1}\times \frac{n}{(n+1)x^n}3^n\right| \\
&= \left|\frac{x}{3}\right|\lim_{n\to\infty} \left|\frac{n(n+2)}{(n+1)^2}\right| \\
&= \frac{|x|}{3}.
\end{split}
\end{equation*}
So if $\frac{|x|}{3} < 1$, then this series will converge. Since $\frac{|x|}{3} < 1 \Longleftrightarrow |x| < 3$, the radius of convergence is $3$, i.e. the series converges for all $x\in (-3,3)$.
Have I gone wrong somewhere or is the question wrong???

REPLY [4 votes]: I don't see anything wrong with what you've done. One way you can basically verify you've done it correctly is to note that, for large $n$, $\frac{n+1}{n} \approx 1$, so the terms basically become $\left(\frac{x}{3}\right)^n$, i.e., a geometric series, which converges for the absolute value of the ratio being less than $1$, i.e., for $\left|\frac{x}{3}\right| \lt 1 \; \implies \; |x| \lt 3$.
Note the question is not "technically" wrong since if the power series converges for all $x \in (-3,3)$, then it'll obviously converge for all $x \in (-1,1)$. However, normally a question would want you to prove a result for the the maximum set of values which converge. As such, it's very reasonable to assume the question didn't intend to ask for $x \in (-1,1)$, and so this is likely a mistake on their part.