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 07:42
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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 07:58
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In my opinion, this question relates to this answer more than to the linked question. – Rócherz Nov 18 '18 at 08:05