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I'd love to understand the behaviour of the sequence $ \frac{(2n)!}{4^n(n!)^2} \text{as } n \to \infty $ the first step would be to simplify this to $ \frac{(2n)(2n-1)(2n-2)\cdots(n+1)}{4^n \cdot n(n-1)(n-2)\cdots 2} $ and then factor out $2$ to get $ \frac{1}{2^n}\cdot\frac{(n)(n-1/2)(n-1)\cdots(n - (n-1)/2)}{n(n-1)(n-2)\cdots 2} $ if I can now get the second term to be strictly larger than $2^n$ then I would be done - but how can I do this ? thanks so much for help!!

P.S. this not a HW question - though it grew out of one where I had to find the radius of convergence for a power series - this is the series evaluated at the end points. If I can show that the above sequence does not converge to $0$ then I know that the power series diverges at the endpoints, this is what I'd love to find out!

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    See [Central binomial coefficient](http://en.wikipedia.org/wiki/Central_binomial_coefficient).2012-12-01

3 Answers 3

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Rewrite the sequence as $ a_n = \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n) }. $

Show that $a_n \le \frac{1}{\sqrt{3n+1}}$ using induction.

Conclude by Sandwich theorem that $\lim_{n\to \infty}a_n = 0$.

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As mentioned before, you can rewrite the sequence as:

$ a_n = \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n) } $

The following inequalities:

$1\cdot3<2^{2}$

$3\cdot5<4^{2}$

$5\cdot7<6^{2}$

$\cdots\cdots\cdots\cdots$

$(2n-3)(2n-1)<(2n-2)^{2}$

$(2n-1)(2n+1)<\left( 2n\right) ^{2}$

multiplied lead to $a_{n}^{2}\cdot(2n+1)<1$, which means that $a_{n}<\dfrac {1}{\sqrt{2n+1}}$. This is enough to guarentee that the limit of $a_{n}$ is $0$.

1

Using $2^n n! = 2 \cdot 4 \cdots (2n)$ you can rewrite the sequence as

$ a_n = \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n) } $

It is rather well known that $\frac{1 / a_n} { \sqrt{n} } = \frac{\frac{2 \cdot 4 \cdots (2n) }{1 \cdot 3 \cdots (2n-1)}}{\sqrt n } \rightarrow \sqrt \pi $

So by taking the inverse of this, $a_n \rightarrow 0$ for $n \rightarrow \infty$.