Generating series of Catalan numbers

Here is a possible line of attack that will give the desired result if we can prove the following (true) lemma

Lemma: For all $n>1$ $$\sum_{k=1}^{n-1} \frac{1}{k^{3/2}(n-k)^{3/2}} \leq \frac{2}{n^{3/2}}$$

Assume $C_k \leq \frac{4^k}{k^{3/2}}$ for $k=1,2,\ldots, n$ then

$$C_{n+1} \leq 4^{n}\left(\frac{2}{n^{3/2}} + \sum_{k=1}^{n-1} \frac{1}{k^{3/2}(n-k)^{3/2}}\right) \leq \frac{4^{n+1}}{(n+1)^{3/2}}$$

by the above lemma and the assumed bound on $C_k$ follows by induction. This result gives

$$\left|\sum_{k=0}^n C_k x^k\right| \leq 1 + \sum_{k=1}^n \frac{(4x)^k}{k^{3/2}}$$

and it follows that the series converges for $|x| < \frac{1}{4}$.

Let $u_N(z) = \sum_{k=0}^N C_k z^k$, a polynomial so no worries about convergence. Use the recurrence relation to show that $z u_N(z)^2 = u_N(z) -1 + O(|z|^{N+1})$. This motivates looking at the equation $z g(z)^2 = g(z) - 1$, which has solution $g(z) = (1 - \sqrt{1-4z})/(2z)$. This (after removing the removable singularity at $0$) is an analytic function in the disk $|z|<1/4$, and its Maclaurin series satisfies the same recurrence and has constant term $1$, so the coefficients are $C_k$. The radius of convergence is $1/4$ because that is the distance from $0$ to the nearest non-removable singularity.