title: Disproving asymptotic lower bound of central binomial coefficient
body: <p>I was recently learning about binomial coefficients and was wondering about how to disprove 2nCn (or the central binomial coefficient) not being lower-bounded by 4^n; in other words:</p>
  <p><img src="https://latex.codecogs.com/gif.latex?%5Cbinom%7B2n%7D%7Bn%7D&space;%5Cneq&space;%5COmega&space;(4%5En)" title="\binom{2n}{n} \neq \Omega (4^n)" /></p>
  <p>Some extremely generous bounds can be easily constructed, such as the following:</p>
  <p><a href="https://i.stack.imgur.com/YuVzx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YuVzx.png" alt="eqn1" /></a></p>
  <p>I sought to prove by contradiction, so to assume:</p>
  
  <p><a href="https://i.stack.imgur.com/C2Hlr.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C2Hlr.gif" alt="eqn2" /></a></p>
  <p>Clearly, c<sub>1</sub> cannot exist, since 1/(2n + 1) approaches 0 as n approaches infinity. It can also be seen that c<sub>2</sub> must reside in (0, 1]. And... I'm stuck. Intuitively, it seems rather obvious that c<sub>2</sub> cannot exist.</p>
  <p>I am aware a similar question has been asked <a href="https://stackoverflow.com/questions/39558086/how-to-prove-binomial-coefficient-is-asymptotic-big-theta-of-two-to-the-power-n">here</a>, but there wasn't really a proof provided. I'm also aware that you could prove the limit of 2nCn/4<sup>n</sup> approaches 0 as n approaches infinity, but I was wondering if there was another way to do so - particularly, by proving that c<sub>2</sub> cannot exist.</p>