Calculating an infinite sum using Newton's generalized binomial theorem Newton's generalized binomial theorem states that $\sum_{k=0}^{\infty} \binom{-2\delta}{k} x^k = (1+x)^{-2\delta}$ for $|x|<1, \delta \in \mathbb{R}$.
Using this, can we calculate the value of the following two sums? It's clear that they converge, but I need help finding the limits. Any help with this would be appreciated.
$$\sum_{k=0}^{\infty} \binom{-2\delta}{k} x^k \frac{1}{k+\delta}$$
$$\sum_{k=0}^{\infty} \binom{-2\delta}{k} x^k \frac{1}{(k+\delta)^2}$$
Again, we may assume $|x|<1, \delta \in \mathbb{R}$.
 A: I suppose $\delta>1$
We have for $k\in\mathbb{N}$,
$\int_0^1 t^{k+\delta-1} dt = \left[\frac{t^{k+\delta}}{k+\delta}\right]_0^1=\frac{1}{k+\delta}$
For $t\in[0,1]$, let $f(t)=\sum_{k=0}^{\infty} \left(\matrix{-2\delta\\k}\right) x^k t^{k+\delta-1}$
We have
$f(t)=t^{\delta-1}\sum_{k=0}^{\infty} \left(\matrix{-2\delta\\k}\right) x^k t^{k}$
$=t^{\delta-1}\sum_{k=0}^{\infty} \left(\matrix{-2\delta\\k}\right) (xt)^k$
$=t^{\delta-1}(1+xt)^{-2\delta}$
The problem is that the integral from 0 to 1 of f is poorly calculated.
Then we exchange the integral with the sum and we obtain an expression of the sum to find.
$\sum_{k=0}^{\infty} \left(\matrix{-2\delta\\k}\right) \frac{x^k}{k+\delta}$
$=\sum_{k=0}^{\infty} \left(\matrix{-2\delta\\k}\right) x^k \int_0^1 t^{k+\delta-1} dt$
$=\int_0^1 f(t) dt$
For the second sum, we can perhaps consider the function g defined by
$g(t)=\sum_{k=0}^{\infty} \left(\matrix{-2\delta\\k}\right) \frac{x^k}{(k+\delta)^2} t^{k+\delta}$
Derivate it
$g'(t)=\sum_{k=0}^{\infty} \left(\matrix{-2\delta\\k}\right) \frac{x^k}{(k+\delta)} t^{k+\delta-1}$
$=t^{\delta-1}\sum_{k=0}^{\infty} \left(\matrix{-2\delta\\k}\right) \frac{(xt)^k}{(k+\delta)}$
Use the previous result, to calculate the sum by replacing x by xt, and finally calculate an anti-derivative of the result...
A: I think that the closed forms would be quite difficult to obtain except using  hypergeometric functions.
Let
$$f_n=\sum_{k=0}^{\infty} \binom{-2d}{k}  \frac{x^k}{(k+d)^n}\qquad \text{with}\qquad n >0$$ What you should get is
$$f_1=\frac 1 {d} \, _2F_1(d,2 d;d+1;-x)$$
$$f_2=\frac 1 {d^2} \, _3F_2(d,d,2 d;d+1,d+1;-x)$$
$$f_3=\frac 1 {d^3} \, _4F_3(d,d,d,2 d;d+1,d+1,d+1;-x)$$
$$f_4=\frac 1 {d^4}\, _5F_4(d,d,d,d,2 d;d+1,d+1,d+1,d+1;-x)$$ showing an obvious pattern
