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Calculate $$\lim_{n\rightarrow +\infty}\binom{2n} n$$ without use Stirling's Formula.

Any suggestions please?

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$$\binom{2n} n = \sum_{k=0}^n \binom{n} k \binom{n}{n-k} \geq \binom{n}1 = n$$

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    $\begingroup$ That's a delightfully weak estimate! +1 $\endgroup$ – mrf Nov 4 '14 at 21:20
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The limit is $+\infty$.

$$\frac {(2n)!}{n!n!} = \frac{2^n n! (2n-1)!!}{n!n!} = 2^n\frac{ (2n-1)!!}{n!}= 2^n\frac{ 1\times 3\times \dots \times (2n-1)}{1\times 2\times \dots \times n} .$$ The first factor goes to infinity quite rapidly and the second factor is quite obviously increasing and positive.

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Simplest way is to use the ratio test: $$ \frac{\binom{2n+2}{n+1}}{\binom{2n}{n}}=\frac{(2n+2)!n!n!}{(2n)!(n+1)!(n+1)!}= \frac{(2n+2)(2n+1)}{(n+1)(n+1)}\to 4, $$ as $n\to\infty$. Hence $$ \binom{2n}{n}\to\infty. $$

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You have:

$${2n\choose n}-{2n\choose n-1}=\frac{1}{n+1}{2n\choose n}$$

These are simply Catalan numbers, by the way. The LHS is an integer, and the RHS is positive, hence, it's a positive integer, and especially it's $\ge1$.

Thus

$${2n\choose n}\geq n+1$$

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By repeated use of the identity $\binom{m}{k}=\frac{m-k+1}{k}\binom{m}{k-1}$, we can see that $\binom{2n}{n} \geq \binom{2n}{k}$ for all $k$.

By the binomial theorem, $4^n=\sum_{k=0}^{2n} \binom{2n}{k}$. As this sum has $2n+1$ terms, the largest of which is $\binom{2n}{n}$, we have $$ \binom{2n}{n} \geq \frac{4^n}{2n+1} $$ which in particular implies that the sequence $\left\{\binom{2n}{n}\right\}$ diverges to $+\infty$.

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