Calculating an infinite sum using Newton's generalized binomial theorem

Question

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 $|x|<1, \delta \in \mathbb{R}$.

Answer 1

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...

Answer 2

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