I was looking for a closed-form expression for the following sum:
$$ \sum_{k=0}^{n-1}(-1)^kC(n-1,k)\frac{x^k}{a+k} $$ where $C(n-1,k)=\frac{(n-1)!}{k!(n-1-k)!}$ is the binomial coefficient, $0<x<1$, and $a>1$. In case there is no closed-form expression for this sum, is there any approximation or upper-bound?
Thank you very much in advance.
Without a loss of generality we can write $$S_n(x)=\sum_{k=0}^{n} (-1)^k {n \choose k} \frac{x^k}{a+k}= \sum_{k=0}^{n} (-1)^k {n \choose k} x^k \int_{0}^{1} t^{a+k-1} dt= \int_{0}^{1} t^{a-1}\sum_{k=0}^{n}(-1)^{k} {n \choose k} (xt)^k$$ $$\implies S_n(x)=\int_{0}^{1}t^{a-1} (1-xt)^n dt = \frac{1}{a}~_2F_1(a,-n;1+a;x)$$ the Gauss Hypergeometric function is written as $$~_2F_1(A,B;C,Z)=1+\frac{AB}{C}\frac{x}{1!}+\frac{A(A+1)B(B+1)}{C(C+1)}\frac{x^2}{2!}+...$$ Serveral interesting properties/approximations of $_2F_1$ can be used from the literature. For example: $$_2F_1(A,B;C;1)=\frac{\Gamma(C)\Gamma(C-A-B)}{\Gamma(C-A) \Gamma(C-B)}$$ will give $$S_n(1)= \frac{\Gamma(1+a) \Gamma(1+n)}{a\Gamma(1+a+n)}=B(a,n+1)=\frac{1}{a{a+n \choose n}}$$ B is the $\beta$ function.