# How to solve this limit (with factorial)? [duplicate]

How to solve this limit??

$$\lim_{k \to \infty} \frac{(2k)!}{2^{2k} (k!)^2}$$

It's a limit, not a series

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• What have you tried? What happens when you try Stirling's approximation? ($n! \sim \sqrt{2\pi n}\left(\frac n e\right)^n$) – MCCCS Nov 18 '18 at 7:42
• Empirical hint. $$\frac{(2k)!}{2^{2k}(k!)^2} =\frac{1}{(2^2)^k} \cdot \frac{(2k)!}{k!(2k-k)!} =\frac{1}{4^k} \binom{2k}{k}.$$ Then look at the numbers in the middle of (the odd rows in) Pascal's Triangle. If you can prove that $$\binom{2(k+1)}{k+1} < 4\binom{2k}{k},$$ then that would imply that $$\binom{2k}{k} <4^{k-1} \binom21.$$ Which then suggests that$\ldots$ – Rócherz Nov 18 '18 at 7:58