The series is in the form of power series, here $x_{0}=0$. We can apply Root Test(We've concluded to use the Root Test in my previous question).
$\displaystyle\sum_{n=1}^{\infty} x^n\left( \dfrac{n}{2n+1} \right)^{2n-1}$. Let $a_{n}= \left | x^n\left( \dfrac{n}{2n+1} \right)^{2n-1} \right |$, it follows that:
$\displaystyle\lim_{n\to\infty} \displaystyle \sqrt[n]{\left | x^n\left( \dfrac{n}{2n+1} \right)^{2n-1} \right |}$ taking limits and we will get $\left |\dfrac{x}{4} \right|$ it must smaller than $1$ to be converge. So Radius of convergence is $4$. Now we must chech end points $x=4$ and $x=-4$. First, I plug $4$ and I used $n^{th}$ term test and I get $\dfrac{1}{2\mathbb{e}}$. So I concluded that the series diverges when $x=4$
Second, I plug $x=-4$ and because I know it doesn't converge absolutely I used Alternating series test(I know because it is same as $x=4$) So I used again $n_{th}$ term test and not surprisingly I get $\dfrac{1}{2\mathbb{e}}$ so again I concluded that the series also diverges at $x=-4$.
All in all, the series diverges iff $x\in \left (\infty, 4 \right]\cup\left [-4, \infty \right)$ and converges iff $x\in\left (-4,4 \right )$. Are there any gaps in my solution? Please verify. Thanks.
The Radius of Convergence of $\sum_{n=1}^{\infty} x^n\left( \frac{n}{2n+1} \right)^{2n-1}$
power seriessolution-verificationsummation
Best Answer
Hint: Observe that the radius of convergence can be calculated using the lim sup given by Hadamard-Cauchy's formula. So if the limit exists, it will squeezed between two limits of ratios given by the ratio test, the lim sup and the lim inf.
Observe that the $1/R = \limsup \limits_{n\to\infty} \sqrt[n]{|c_n|}$,
where $c_n$ is giving by $\sum_{i = 0}^{\infty} c_nx^n$.