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

marked as duplicate by gimusi limits StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 18 '18 at 7:44

• 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