# Thread: Convergence Problem

1. ## Convergence Problem

2. 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.

3. Originally Posted by szpengchao
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

4. Originally Posted by flyingsquirrel
What link is there between "$\displaystyle a_n\to_0$ for $\displaystyle n\to\infty$" and "$\displaystyle \sum a_n$ diverges" ?
Yeah I saw that...I changed it before you posted....I meant convergent by alternating series test

5. ## check

please check if my proof for the first part is right

6. Originally Posted by szpengchao
please check if my proof for the first part is right
where did that come from?

7. ## first step

the first step is wallis product

8. Originally Posted by szpengchao
$\displaystyle \lim_{n\to\infty}\frac{(2n)!}{(n!)^24^n}$

Letting L equal that limit we get

Take the natural logs of both sides

and note taht $\displaystyle \ln((2n)!)=\sum_{m=1}^{2n}\ln(m)$

9. Originally Posted by szpengchao
please check if my proof for the first part is right
If you're allowed to use the Wallis product, (Is it in the exercise ?) it's OK.

10. ## good good

just want to check if there are some wrong steps...

ok...cheers!

so for the last part, you did ratio test and try to find the radius of convergence, right?

then the result comes straight from it?

11. Originally Posted by szpengchao
just want to check if there are some wrong steps...

ok...cheers!

so for the last part, you did ratio test and try to find the radius of convergence, right?

then the result comes straight from it?
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 ____

12. ## no idea

i have no idea to show when x=1/4

please give some hint

13. Originally Posted by szpengchao
i have no idea to show when x=1/4
Originally Posted by flyingsquirrel
• $\displaystyle x=\frac{1}{4}$ : Are you allowed to use Stirling's approximation ? (for $\displaystyle n\to\infty$, $\displaystyle n!\sim\sqrt{2\pi n}\left(\frac{n}{\mathrm{e}}\right)^n$)
...

14. ## yeah

yeah...i got as n goes to infinity, that is equivalent to 1/sqrt( pi * n)

so, what's next?

15. 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.

Page 1 of 2 12 Last