Evaluate the infinite series $$\sum_{n=0}^\infty \frac{\binom{2n+1}{n-p}}{(4^n)(2n+1)(1.25)^{2n+1}}$$
On a calculator, for $p=0$ this tends to $1$. For $p=1$ it tends to $\frac{1}{12}$. For all values of $p$ I checked the result was a reciprocal of a natural number. How would you calculate this without a calculator for single values of $p$ and for any $p$ in general (natural numbers).
Hint. Note that $$\begin{align}\sum_{n=0}^\infty \frac{\binom{2n+1}{n-p}}{(4^n)(2n+1)(1.25)^{2n+1}}&= 2\sum_{n=0}^\infty \frac{\binom{2n+1}{n+p+1}}{(2n+1)(2.5)^{2n+1}}\\ &=\frac{4}{5}\sum_{n=0}^\infty \frac{\binom{2n}{n+p}x^n}{n+p+1} \end{align}$$ with $x=4/25$. For $p=0$, recall that, for $|x|<1/4$, $$\sum_{n=0}^{\infty} \frac{\binom{2n}{n}x^n}{n+1} = \frac{1-\sqrt{1-4x}}{2x}=\frac{2}{1+\sqrt{1-4x}}$$ (see the g.f.of the Catalan Numbers). Extend this identity and show that for non-negative integer $p$, $$\sum_{n=0}^{\infty} \frac{\binom{2n}{n+p}x^n}{n+p+1} = \left(\frac{2}{1+\sqrt{1-4x}}\right)^{2p+1}\cdot \frac{x^p}{2p+1}$$