Radius of convergence of series $\sum_{n=0}^{\infty}\frac{(x-1)^{2n}}{2^n n^3}$

Question

My Attempt : $$a_{2n}=\frac{1}{2^{n}n^{3}}$$

But Root Test gives $$a_{2n}^{1/2n}=\frac{1}{\sqrt{2}}$$ and Ratio Test gives $$\frac{a_{n+1}}{a_n}=\frac{1}{2}$$ So What is ROC $\mathbf{✓2 \;or\; 2}$

Answer 1

You can define $$ a_n : = \frac{{(x - 1)^{2n} }}{{2^n n^3 }}. $$ Then $$ \sqrt[n]{{a_n }} = \frac{{(x - 1)^2 }}{{2n^{3/n} }} \to \frac{{(x - 1)^2 }}{2} $$ and $$ \frac{{a_{n + 1} }}{{a_n }} = \frac{{(x - 1)^2 n^3 }}{{2(n + 1)^3 }} \to \frac{{(x - 1)^2 }}{2}. $$ Thus, by either the root or the ratio test, the series is convergent when $$ \frac{{(x - 1)^2 }}{2} < 1 \Leftrightarrow \left| {x - 1} \right| < \sqrt 2 . $$

Answer 2

I suppose that the power series is given by

$$ \sum_{n=0}^{\infty}\frac{(x-1)^{2n}}{2^n 3^n}.$$

The root test leads to

$$\frac{(x-1)^2}{6}<1,$$

hence

$$|x-1| < \sqrt{6}.$$

Answer 3

$\sum\dfrac{|x-1|^{2n}}{(2\cdot3)^n}=$

$\sum \left (\dfrac{|x-1|}{√6}\right )^{2n}$.

Converges for $\dfrac{|x-1|}{√6}<1$.

Answer 4

The ratio formula cannot be applied here since

• $\sum_{n=1}^{\infty}\frac{(x-1)^{2n}}{2^n n^3} = \sum_{k=1}^{\infty}a_k(x-1)^k$ with $$a_{2l+1}=0 \text{ and } a_{2l}=\frac 1{2^ll^3} \text{ for } l \in \mathbb{N} $$

So, the quotient $\frac{a_{k+1}}{a_k}$ is either $0$ or undefined.

But with the root formula there is no problem:

$$R = \frac 1{\limsup \sqrt[n]{a_n}} = \frac 1{\lim_{l\to\infty} \sqrt[2l]{a_{2l}}} = \lim_{l\to\infty} (\sqrt 2 \sqrt[2l]{l^3})=\sqrt{2}$$

Answer 5

To apply the ratio test for the series, we do this: $$ \frac{(x-1)^{2n+2}/2^{n+1} (n+1)^3}{(x-1)^{2n}/2^n n^3} =\frac{(x-1)^2}{2}\;\cdot\;\frac{n^3}{(n+1)^3} \to \frac{(x-1)^2}{2} $$ Thus we have convergence if $$ \left|\frac{(x-1)^2}{2}\right| < 1 \Longleftrightarrow |(x-1)^2| < 2 \Longleftrightarrow |x-1| < \sqrt2 $$ Similarly we have divergence if $|x-1| > \sqrt{2}$. As usual, the ratio text is inconclusive when $|x-1| = \sqrt2$.