Let $X$ be a random variable and it is known that the mgf of $X$ exists.
If the $k$th moment is given by $m_k=\mathbb E[X^k]=\frac{(2k+1)!}{k!2^k}$ for $k=0, 1, ...$
Problem: Find the mgf of $X$.
My attempt: The mgf of $X$ is $M_X (t)= \sum_{k=0}^\infty \frac{m_k}{k!}t^k=\sum_{k=0}^\infty \frac{(2k+1)!}{(k!)^22^k}t^k$.
However, I have no idea how to proceed further. What should I do?
$$\sum_{k=0}^{\infty}\binom{2k}{k}x^k = (1-4x)^{-1/2}$$
Put $x = \frac{t}{2}$,
$$\sum_{k=0}^{\infty}\binom{2k}{k}\left(\frac{t}{2}\right)^k = (1-2t)^{-1/2}$$
Take derivative wrt $t$,
$$\sum_{k=0}^{\infty}\binom{2k}{k}\frac{k}{2}\left(\frac{t}{2}\right)^{k-1} = (1-2t)^{-3/2}$$
multiply with $t$,
$$\sum_{k=0}^{\infty}\binom{2k}{k}k\left(\frac{t}{2}\right)^{k} = t(1-2t)^{-3/2}$$
Finally,
$$\sum_{k=0}^{\infty}\binom{2k}{k}(2k+1)\left(\frac{t}{2}\right)^{k} = 2t(1-2t)^{-3/2} + (1-2t)^{-1/2} = (1-2t)^{-3/2}$$
So, the distribution is,
$$gamma\left(\frac{3}{2}, 2\right)$$
Check this question for the proof of first equation.