I am having difficulty proving the following result
\begin{align}
\sum_{n=0}^{\infty} \binom{2n}{n} \frac{x^{n}}{n!} = e^{2x} \ I_{0} (2x) \text{,}
\end{align}
where
\begin{align}
I_{0}(y) = \sum_{n=0}^{\infty} \frac{\Big( \frac{y}{2} \Big)^{2n}}{n! \ n!}
\end{align}
is a modified Bessel function of the first kind with zero order.
I approached it using
\begin{align}
e^{2x} \ I_{0} (2x) = \bigg( \sum_{n=0}^{\infty} \frac{(2x)^{n}}{n!}\bigg) \ \bigg( \sum_{k=0}^{\infty} \frac{ x^{2k}}{k! \ k!} \bigg) \text{,}
\end{align}
but I have had no luck.
Best Answer
Extracting the coefficient we seek to show
$${2n\choose n} \frac{1}{n!} = [z^n] \sum_{k\ge 0} \frac{2^k}{k!} z^k \sum_{k\ge 0} \frac{1}{k! \times k!} z^{2k}.$$
This is
$${2n\choose n} \frac{1}{n!} = \sum_{q=0}^{\lfloor n/2 \rfloor} \frac{1}{q! \times q!} \frac{2^{n-2q}}{(n-2q)!}.$$
or
$${2n\choose n} = \sum_{q=0}^{\lfloor n/2 \rfloor} \frac{1}{q! \times q!} \frac{2^{n-2q} \times n!}{(n-2q)!}.$$
or
$${2n\choose n} = \sum_{q=0}^{\lfloor n/2 \rfloor} \frac{(n-q)!}{q! \times q!} \frac{2^{n-2q} \times n!}{(n-q)! \times (n-2q)!}.$$
In terms of binomial coefficients
$${2n\choose n} = \sum_{q=0}^{\lfloor n/2 \rfloor} {n\choose q} {n-q\choose q} 2^{n-2q}.$$
The RHS is
$$\sum_{q=0}^{\lfloor n/2 \rfloor} {n\choose q} {n-q\choose n-2q} 2^{n-2q} \\ = 2^n [z^n] (1+z)^n \sum_{q=0}^{\lfloor n/2 \rfloor} {n\choose q} 2^{-2q} (1+z)^{-q} z^{2q}.$$
The coefficient extractor combined with the $z^{2q}$ factor enforces the upper limit and we may write
$$2^n [z^n] (1+z)^n \sum_{q\ge 0} {n\choose q} 2^{-2q} (1+z)^{-q} z^{2q} \\ = 2^n [z^n] (1+z)^n \left(1+\frac{z^2}{4(1+z)}\right)^n \\ = 2^n [z^n] (1+z)^n \frac{(4+4z+z^2)^n}{4^n (1+z)^n} \\ = 2^{-n} [z^n] (z+2)^{2n} \\ = 2^{-n} {2n\choose n} 2^n = {2n\choose n}.$$
This is the claim.