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?
 A: 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:
$$G(t) = \frac{1}{(1-t)^2}\frac{1+\xi t}{(1-t)^\xi}+\frac{\xi(1-t)}{(1-t)^2}\frac{1}{(1-t)^\xi}=\frac{1+\xi}{(1-t)^{2+\xi}}.$$
A: 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,
$$G(t|m)\approx \frac{1}{m!}\sum_{n=0}^\infty \frac{t^n (n+m+1)!}{n!}=\frac{1}{m!}S_m=\frac{1}{m!}\frac{\mathrm{d}^{m+1}}{\mathrm{d}t^{m+1}}\frac{1}{1-t}=\frac{m+1}{(1-t)^{m+2}}.$$
A: Both of the other answers are nice, but I think this is what I was looking for, so I'm answering my question:
It can be easily shown with induction that
$$B(n+1,\xi+1)=\frac{n!}{(\xi+1)\cdots(\xi+n+1)}$$ for all $n$. All it takes is an integration by parts on the formula of the Beta function:
Now we consider the function $\psi(t)=(\xi+1)(1-t)^{-(\xi+2)}$. Again, by induction it is easily proved that
$$\psi^{(n)}(t)=(\xi+1)\cdots(\xi+n+1)(1-t)^{-(\xi+n+2)}$$
So $\psi^{(n)}(0)/n!=1/B(n+1,\xi+1)$. This shows that our series is exactly the power series for $\psi(t)$, i.e.
$$G(t)=\sum_{n=0}^\infty\frac{\psi^{(n)}(0)}{n!}t^n=\psi(t)=\frac{\xi+1}{(1-t)^{\xi+2}} $$
