A curious limit

First we see the main problem i.e. for the limit:

$\begin{aligned} \displaystyle \lim_{n \to \infty} \left| \dfrac{a_{n+1}}{a_n} \right| &= \lim_{n \to \infty} \left| \dfrac{\dbinom{2n}{n}}{\dbinom{2n+2}{n+1}} \right| \\ &= \lim_{n \to \infty} \dfrac{{(2n)!}/{(n!)^2}}{{(2n+2)!}/{{((n+1)!)}^2}} \\ &= \lim_{n \to \infty} \dfrac{n+1}{2 (2n+1)} \\ &= \lim_{n \to \infty} \dfrac{1+\frac 1n}{2 \left( 2 + \frac 1n \right)} \\ &= \dfrac{1}{4} \\ &= \boxed{0.25} \end{aligned}$

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Now, heading onto the more challenging bonus :

Since by ratio test we have already shown that $\displaystyle \lim_{n \to \infty} \left| \dfrac{a_{n+1}}{a_n} \right| = 0.25 < 1$ , therefore the series converges. Its sum is calculated as follows

$\begin{aligned} \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{\dbinom{2n}{n}} &= \sum_{n=1}^{\infty} \dfrac{(n!) \cdot (n!)}{(2n)!} \\ &= \sum_{n=1}^{\infty} \dfrac{\Gamma (n+1) \Gamma (n+1)}{\Gamma (2n+1)} \\ &= \sum_{n=1}^{\infty} (2n+1) \cdot \dfrac{\Gamma (n+1) \Gamma (n+1)}{\Gamma (2n+2)} \\ &= \sum_{n=1}^{\infty} (2n+1) \cdot B(n+1,n+1) \\ &= \sum_{n=1}^{\infty} (2n+1) \cdot \int_0^1 t^n (1-t)^n \, dt \\ &= 2 \sum_{n=1}^{\infty} \int_0^1 n t^n (1-t)^n \, dt + \sum_{n=1}^{\infty} \int_0^1 t^n (1-t)^n \, dt \\ &= 2 \int_0^1 \sum_{n=1}^{\infty} n t^n (1-t)^n \, dt + \int_0^1 \sum_{n=1}^{\infty} t^n (1-t)^n \, dt \qquad \qquad \qquad \small{\color{#3D99F6}\text{interchanging order of summation and integral by Fubini's theorem}} \\ &= 2 \int_0^1 \dfrac{t(t-1)}{(t(t-1)+1)^2} \, dt + \int_0^1 \dfrac{t(t-1)}{(t(t-1)+1)} \, dt \\ &= 2 {\left[ \dfrac{1-2t}{3(t^2-t+1)} + \dfrac{2}{3 \sqrt{3}} \arctan \left( \dfrac{2t-1}{\sqrt{3}} \right) \right]}_0^1 + {\left[ t - \dfrac{2}{\sqrt{3}} \arctan \left( \dfrac{2t-1}{\sqrt{3}} \right) \right]}_0^1 \\ &= 2 \left[ - \dfrac{2}{27} (\sqrt{3} \pi - 9) \right] + \left[ \dfrac{2 \pi}{3 \sqrt{3}} - 1 \right] \\ &= - \dfrac{4 \sqrt{3} \pi}{27} + \dfrac{4}{3} + \dfrac{2 \pi}{3 \sqrt{3}} - 1 \\ &= \boxed{\dfrac{1}{27} (2 \sqrt{3} \pi + 9)} \end{aligned}$

$\large \displaystyle \lim_{n \rightarrow \infty} \left | \frac{a_{n+1}}{a_n} \right |$

$\large \displaystyle = \lim_{n \rightarrow \infty} \left | \frac{(n+1)!(n+1)!}{(2n+2)!} \cdot \frac{(2n)!}{n!n!} \right |$

$\large \displaystyle = \lim_{n \rightarrow \infty} \left | \frac{(n+1)(n+1)}{(2n+2)(2n+1)} \right |$

$\large \displaystyle = \lim_{n \rightarrow \infty} \left | \frac{n^2 \cdot (1 + \frac2n + \frac1{n^2}) }{n^2 \cdot (4 + \frac6n + \frac2{n^2})} \right |$

$\color{#3D99F6} \boxed{\large \displaystyle = \frac14 = 0.25}$

The limit being $\tfrac14$ is easy enough. As for the series, look at the solutions given here .