Study the extremes of the radio convergence of $$\sum_{n=1}^\infty{\frac n{2^n(2n-1)}{(x-1)}^n}\,.$$
I know that, by the radio test, the convergence of this series is for $x\in(-1,3)$. So let's study when:
- $\boxed{x=-1}:\displaystyle\sum_{n=1}^\infty{\frac n{2^n(2n-1)}{(-1-1)}^n} =\displaystyle\sum_{n=1}^\infty{\frac n{2^n(2n-1)}{(-1)}^n2^n} =\displaystyle\sum_{n=1}^\infty{{(-1)}^n\underbrace{\frac{n}{2n-1}}_{a_n}}\,, $
so now I will apply Leibniz Rule for $a_n$:
I) $a_n\geq0$?: Yes;
II) $\displaystyle\lim_{n\to\infty}{a_n}=0$? NO, because $$\displaystyle\lim_{n\to\infty}{a_n}=\displaystyle\lim_{n\to\infty}{\frac{n}{2n-1}}=\frac 12\neq 0,$$ therefore $$\displaystyle\sum_{n=1}^\infty{{(-1)}^n\frac{n}{2n-1}}\qquad\text{diverges}.$$
Similarly, when
- $\boxed{x=3}:\displaystyle\sum_{n=1}^\infty{\frac n{2^n(2n-1)}{(3-1)}^n} =\displaystyle\sum_{n=1}^\infty{\frac n{2^n(2n-1)}2^n} =\displaystyle\sum_{n=1}^\infty{\frac{n}{2n-1}}\qquad\text{diverges}, $
therefore $$\boxed{\text{the series absolutely converges when}\qquad x\in(-1,3)}\quad\text{not }x\in[-1,3].$$
Is that ok?
Thank you!!
Yes of course it’s the right result but more simply in both case it suffices to note that we have $$a_n \not \to 0$$ which is a necessary condition for convergence (we don’t need to refer to Leibniz).