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?

Well, I think I can give an approximate answer. Let $m := \operatorname{nint}(\xi)$, the nearest integer function. Then we can rewrite:
$$G(t|m)\approx\sum_{n=0}^\infty \frac{t^n}{B(n+1,m+1)}$$
Using the property of the beta function that $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$, and the fact that $\Gamma(N+1)=N!$ for $N\in\mathbb{N}$,
$$G(t|m)\approx \frac{1}{m!}\sum_{n=0}^\infty \frac{t^n (n+m+1)!}{n!}$$
Following from @Sangchul Lee's work here, to derive the sum, we can define
$$S_m=\sum_{n=0}^\infty\frac{t^n(n+m+1)!}{n!}$$
We notice that
$$S_{-1}=1+t+t^2+t^3+...=\frac{1}{1-t}$$
Furthermore,
$$\frac{\mathrm{d}}{\mathrm{d}t}S_m=\frac{\mathrm{d}}{\mathrm{d}t}\left(
(m+1)!+t(m+2)!+t^2\frac{(m+3)!}{2!}+t^3\frac{(m+4)!}{3!}+...\right)$$
$$=\left((m+2)!+t(m+3)!+t^2\frac{(m+4)!}{2!}+...\right)=S_{m+1}$$
Therefore
$$S_m=\frac{\mathrm{d}^{m+1}}{\mathrm{d}t^{m+1}}\frac{1}{1-t}$$
$$\frac{\mathrm{d}}{\mathrm{d}t} (1-t)^{-1}=-1(1-t)^{-2}(-1)=\frac{1}{(1-t)^2}$$
$$\frac{\mathrm{d}}{\mathrm{d}t} (1-t)^{-2}=(-2)(1-t)^{-3}(-1)=\frac{2}{(1-t)^3}$$
$$\frac{\mathrm{d}}{\mathrm{d}t}2(1-t)^{-3}=\frac{2\cdot 3}{(1-t)^4}$$
And so on. In general,
$$\frac{\mathrm{d}^n}{\mathrm{d}t^n}\frac{1}{1-t}=\frac{n!}{(1-t)^{n+1}}$$
Therefore,
$$G(t|m)\approx \frac{1}{m!}\sum_{n=0}^\infty \frac{t^n (n+m+1)!}{n!}=\frac{1}{m!}S_m=\frac{1}{m!}\frac{\mathrm{d}^{m+1}}{\mathrm{d}t^{m+1}}\frac{1}{1-t}=\frac{m+1}{(1-t)^{m+2}}.$$
