Hi
ii) follow the hint which is given...
iii)
- $\displaystyle x=\frac{1}{4}$ : Are you allowed to use Stirling's approximation ? (for $\displaystyle n\to\infty$, $\displaystyle n!\sim\sqrt{2\pi n}\left(\frac{n}{\mathrm{e}}\right)^n$)
- $\displaystyle x=-\frac{1}{4}$ : Think about the alternating series test.
The second one is asking for
$\displaystyle \sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^2}x^n$
First you do rato test
$\displaystyle \lim_{n\to\infty}\bigg|\frac{x^{n+1}(2n+2)!}{((n+1 )!)^2}\cdot\frac{(n!)^2}{(2n)!x^n}\bigg|=\infty$
Since simplifying we would have $\displaystyle \lim_{n\to\infty}\bigg|\frac{x\cdot{x^{n}}(2n+2)(2 n+1)(2n)!}{(n+1)^2(n!)^2}\cdot\frac{(n!)^2}{(2n)!x ^b}\bigg|=\lim_{n\to\infty}|4x|<1\Rightarrow{x<\fr ac{1}{4}\text{or}x>\frac{-1}{4}}$
Now we can see that $\displaystyle \sum_{n=0}^{\infty}\frac{(2n)!(-1)^n}{(n!)^24^n}$
Converges since $\displaystyle \lim_{n\to\infty}\bigg|\frac{(2n)!(-1)^n}{(n!)^24^n}\bigg|=\lim_{n\to\infty}\frac{(2n) !}{(n!)^24^n}=0$ and $\displaystyle a_{n+1}<a_n$
Now for the other test use stirlings approximation
I showed you the first part in case you needed the interval of convergence
Yes I found the radius of convergence and showed that $\displaystyle \sum{a_n}$
is convergent $\displaystyle \forall{x}\in\bigg(\frac{-1}{4},\frac{1}{4}\bigg)$
Then we msut check the endpoints...through the alternating series test we can see that $\displaystyle x=\frac{-1}{4}$ makes the series converge
and I leave it up to you to show that the other value makes the series ____
As for $\displaystyle n\to\infty$, $\displaystyle \left|\binom{2n}{n}\frac{1}{4^n}\right| \sim \frac{1}{\sqrt{\pi n}}$ and $\displaystyle \sum_{n\geq 1}\frac{1}{\sqrt{\pi n}}$ diverges, (p-series with $\displaystyle p=\frac{1}{2}\leq 1$) the series $\displaystyle \sum_{n\geq 1}\binom{2n}{n}\frac{1}{4^n}$ diverges too. It's a theorem but I don't know how it is called in English.