# Thread: Find the closed-form of the series.

1. ## Find the closed-form of the series.

Prove that

$\displaystyle \sum_{n=0}^{\infty}{{2n}\choose n} x^n$ diverges for $\displaystyle |x| = 1/4$.

Also find the closed-form of the series for x in the radius of convergence.

2. As for $\displaystyle |x| < \frac{1}{4}$ the closed form can be obtained from applying the binomial series to $\displaystyle (1-4x)^{-1/2}$.

3. Originally Posted by mathwizard
Prove that

$\displaystyle \sum_{n=0}^{\infty}{{2n}\choose n} x^n$ diverges for $\displaystyle |x| = 1/4$.

Also find the closed-form of the series for x in the radius of convergence.
$\displaystyle \lim_{n\to\infty}\frac{\sqrt{4\pi{n}}(n^n)^2(e^{-n})^24^n}{2\pi{n}(n^n)^2(e^{-n})^24^n}=0$

Therefore when $\displaystyle x=\frac{-1}{4}$

The series converges

4. Originally Posted by Mathstud28
$\displaystyle \lim_{n\to\infty}\frac{\sqrt{4\pi{n}}(n^n)^2(e^{-n})^24^n}{2\pi{n}(n^n)^2(e^{-n})^24^n}=0$

Therefore when $\displaystyle x=\frac{-1}{4}$

The series converges
Huh? What test are you using to determine the convergence (are you using Stirling's approximation for $\displaystyle n!$?)?

The ratio test seems to yield an inconclusive result (i.e., the limit is 1).

5. For $\displaystyle x=1/4$ I think you can use Abel's theorem. It would mean the series is continous at $\displaystyle x=1/4$ which is a problem because it is equal to $\displaystyle (1-4x)^{-1/2}$ when $\displaystyle |x|<1/4$ and therefore not defined at $\displaystyle x=1/4$.

6. Originally Posted by mathwizard
Huh? What test are you using to determine the convergence (are you using Stirling's approximation for $\displaystyle n!$?)?

The ratio test seems to yield an inconclusive result (i.e., the limit is 1).
Correct me if I am wrong but,

Let $\displaystyle a_n=\frac{(2n)!}{4^n(n!)^2}$

Well

$\displaystyle \exists{N}\backepsilon\forall{n>N}\text{ }a_{n+1}<a_n$

and $\displaystyle \lim_{n\to\infty}a_n=0$

Thus the series converges by the alternating series test