Series with coefficients coming from beta function I recently came across the series
$$G(t):=\sum_{n=0}^\infty\frac{t^{n}}{B(n+1,\xi+1)}$$
defined for $t\in(0,1)$ and $\xi+1>0$. I am trying to get a better-looking form for this sum, something more "usable" in general, but I don't know how to handle the beta function. I am pretty sure that this is the power series for $g(t)=\frac{1}{(1-t)^{2+\xi}}$, but I can't prove it. Right now I have only noticed that for $\xi=0$ we get $G(t)=\sum_{n\geq0}(n+1)t^{n}=\frac{1}{(1-t)^2}$, the derivative of the geometric series. Any ideas on how to deal with this sum?

Here is the general non-integer solution (leading to the same solution as @K.dafaoite integer solution). From the paper by E.Stade The reciprocal of the beta function we see that if $n+\xi+1>0$:
$$\frac{1}{B(n+1, \xi+1)}=\frac{n+\xi+1}{2\pi i}\int_{|u|=1}\left(1+\frac{1}{u}\right)^n(1+u)^\xi\frac{du}{u}$$
where the integral is taken counterclockwise in the complex plane.
Therefore, $$G(t)=\sum_{n=0}^\infty t^n\cdot\frac{n+\xi+1}{2\pi i}\int_{|u|=1}\left(1+\frac{1}{u}\right)^n(1+u)^\xi\frac{du}{u}$$
Changing order of summation and integration we get
$$G(t)=\frac{1}{2\pi i}\int_{|u|=1}\sum_{n=0}^\infty(n+\xi+1)\left(1+\frac{1}{u}\right)^nt^n(1+u)^\xi\frac{du}{u}=$$
$$=\frac{1}{2\pi i}\int_{|u|=1}\sum_{n=0}^\infty(n+1)\left(1+\frac{1}{u}\right)^nt^n(1+u)^\xi\frac{du}{u}+\frac{\xi}{2\pi i}\int_{|u|=1}\sum_{n=0}^\infty\left(1+\frac{1}{u}\right)^nt^n(1+u)^\xi\frac{du}{u}=$$
$$=\frac{1}{2\pi i}\int_{|u|=1}\frac{(1+u)^\xi}{\left(1-\left(1+\frac{1}{u}\right)t\right)^2}\frac{du}{u}+\frac{\xi}{2\pi i}\int_{|u|=1}\frac{(1+u)^\xi}{\left(1-\left(1+\frac{1}{u}\right)t\right)}\frac{du}{u}=$$
which can be slightly simplified, leading to
$$G(t) = \frac{1}{(1-t)^2}\frac{1}{2\pi i}\int_{|u|=1}\frac{u(1+u)^\xi}{\left(u-\frac{t}{1-t}\right)^2}du+\frac{\xi}{1-t}\frac{1}{2\pi i}\int_{|u|=1}\frac{(1+u)^\xi}{(u-\frac{t}{1-t})}du.$$
Let's compute the second integral first. Using the residue theorem,
$$\frac{1}{2\pi i}\int_{|u|=1}\frac{(1+u)^\xi}{(u-\frac{t}{1-t})}du={\tt{Res}}_{u=t/(1-t)}f(u)=\frac{1}{(1-t)^\xi}.$$
Similarily, the first integral (having double singularity) leads to
$$\frac{1}{2\pi i}\int_{|u|=1}\frac{u(1+u)^\xi}{\left(u-\frac{t}{1-t}\right)^2}du = \left.\frac{d}{du}\right|_{u=1/(1-t)}u(u+1)^\xi=\frac{1+\xi t}{(1-t)^\xi}$$
Combinig the integrals leads to:
$$G(t) = \frac{1}{(1-t)^2}\frac{1+\xi t}{(1-t)^\xi}+\frac{\xi(1-t)}{(1-t)^2}\frac{1}{(1-t)^\xi}=\frac{1+\xi}{(1-t)^{2+\xi}}.$$
