When I was playing with Binomial coefficients. I got an interesting problem. It is very nice formula. The problem is
$$\large \sum_{n=0}^{\infty} \dfrac{1}{\binom{2n}{n}}$$
where $\binom{2n}{n}$ represents the number of ways of selecting $n$ different things from $2n$ different things. I will provide the solution below.
Before we start the solution we shall learn about two functions.
1) Gamma Function: It is represented by $\Gamma(x)$ and it is given by the expression
$$\Gamma(x) = \int_{0}^{\infty} e^{-t} t^{x-1} dt$$
It has some special properties. I shall mention those properties but not their proofs.
$\Gamma(x+1) = x \Gamma(x)$. This can be proved by integration by parts.
$\Gamma(1) = 1$.
Gamma function is undefined for non-positive integers.
For every non-negative integer $n$, $\Gamma(n+1) = n!$
2) Beta Function: It is represented by $B(x,y)$ and it is given by the expression
$$B(x,y) = \int_{0}^{1} t^{x-1}(1-t)^{y-1} dt$$
Relation between Gamma and Beta function:
$$B(x,y) = \dfrac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$$
Now let us move on to evaluating our summation. You should need to know another thing i.e. the value of $\binom{2n}{n} = \dfrac{(2n)!}{(n)! (n)!}$
First we shall convert the factorial into Beta function.
$$\begin{align*} \binom{2n}{n} &= \dfrac{(2n)!}{(n)! (n)!} \\ &= \dfrac{\Gamma(2n+1)}{\Gamma(n+1) \Gamma(n+1)} \\ &= \dfrac{1}{2n+1} \dfrac{(2n+1)\Gamma(2n+1)}{\Gamma(n+1) \Gamma(n+1)} \\ &= \dfrac{1}{2n+1} \dfrac{\Gamma(2n+2)}{\Gamma(n+1) \Gamma(n+1)} \\ \dfrac{1}{\binom{2n}{n}} &= (2n+1) \dfrac{\Gamma(n+1) \Gamma(n+1)}{\Gamma(2n+2)}\\ &= (2n+1) B(n+1,n+1) \end{align*}$$
$$\begin{align*} \sum_{n=0}^{\infty} \dfrac{1}{\binom{2n}{n}} &= \sum_{n=0}^{\infty} (2n+1) \int_{0}^{1} t^n(1-t)^n dt \\ &= \int_{0}^{1} \sum_{n=0}^{\infty} (2n+1)(t(1-t))^n dt \end{align*}$$
Here, we have to evaluate an infinite summation i.e. $1+3x+5x^2 + 7x^3 + \ldots$.
$$\begin{align*} S &=1+3x+5x^2 + 7x^3 + \ldots \\ &= (1+ 2x+2x^2 + 2x^3 + \ldots ) + (x + 3x^2 + 5x^3 + \ldots ) \\ &= 2(1+x+x^2 +x^3 +\ldots) +xS -1 \\ S(1-x) &= \dfrac{2}{1-x} - 1 \\ S &= \dfrac{2}{(1-x)^2} - \dfrac{1}{1-x} \end{align*}$$
So, $$\sum_{n=0}^{\infty} (2n+1)x^n = \dfrac{2}{(1-x)^2} - \dfrac{1}{1-x} \\ \implies \sum_{n=0}^{\infty} (2n+1)(t(1-t))^n = \dfrac{2}{(1-t+t^2)^2} - \dfrac{1}{1-t+t^2} $$
$$\int_{0}^{1} \sum_{n=0}^{\infty} (2n+1)(t(1-t))^n dt = \int_{0}^{1} \dfrac{2}{(1-t+t^2)^2} dt - \int_{0}^{1} \dfrac{1}{1-t+t^2} dt $$
Now we have to evaluate the above integrals.
Let $ I=\int_{0}^{1} \dfrac{1}{1-t+t^2} dt$ and $J = \int_{0}^{1} \dfrac{1}{(1-t+t^2)^2} dt$
First Integral (I):
$$\begin{align*} I &= \int_{0}^{1} \dfrac{1}{1-t+t^2} dt \\ &= \int_{0}^{1} \dfrac{1}{(t-1/2)^2 + (\sqrt{3}/2)} dt \quad \text{Take the substitution } y= t- 1/2 \\ &= \int_{-1/2}^{1/2} \dfrac{1}{y^2 + (\sqrt{3}/2)^2} dy \\ &= \dfrac{2}{\sqrt{3}} \arctan \left(2y/\sqrt{3} \right) \Bigr|_{-1/2}^{1/2} \\ &= \dfrac{2}{\sqrt{3}} \dfrac{\pi}{3} \\ &=\dfrac{2 \pi}{3 \sqrt{3}} \end{align*}$$
Second Integral (J):
$$\begin{align*} J &= \int_{0}^{1} \dfrac{1}{(1-t+t^2)^2} dt \\ &= \int_{0}^{1} \dfrac{1}{((t-1/2)^2 + (\sqrt{3}/2)^2)^2} dt \quad \text{Take the substitution } y= t-1/2 \\ &= \int_{-1/2}^{1/2} \dfrac{1}{(y^2 + (\sqrt{3}/2)^2)^2} dy \quad \text{Take the substitution } y= \dfrac{\sqrt{3}}{2} \tan \theta \\ &= \dfrac{8}{3 \sqrt{3}} \int_{-\pi /6}^{\pi /6} \cos^2 \theta d \theta \\ &= \dfrac{4}{3 \sqrt{3}} \int_{\pi /6}^{\pi /6} (1+ \cos (2 \theta)) d \theta \\ &= \dfrac{4}{3 \sqrt{3}} \left( \dfrac{\pi}{3} + \dfrac{\sqrt{3}}{2} \right) \end{align*}$$
Now,
$$\begin{align*} \sum_{n=0}^{\infty} \dfrac{1}{\binom{2n}{n}} &= 2J - I \\ &= \boxed{\large{\dfrac{2 \pi}{9 \sqrt{3}}} + \dfrac{4}{3}} \end{align*}$$