A general expression for the summation of series with binomial coefficients : $\sum_{k=0}^{n}(-1)^k{{n}\choose{k}} \frac{1}{k+n}$

Question

I try to derive a general expression for the following summation of series: $$S_n=\sum_{k=0}^{n}(-1)^k{{n}\choose{k}} \frac{1}{k+n},$$ where $n$ is a positive integer.

I have computed it manually: $$S_1=\frac{1}{2},\hspace{0.2cm}S_2=\frac{1}{3\times4},\hspace{0.2cm}S_3=\frac{1}{2\times5\times6},\hspace{0.2cm}S_4=\frac{1}{5\times7\times8}.$$ But I couldn't found any regularity to set a hypothesis for $S_n$.

Could someone help me? Thanks a lot!

Answer 1

Hint

$$ x^{n-1}(1-x)^n = \sum_{k=0}^n (-1)^k{{n}\choose{k}}x^{n+k-1} $$

$$ \int \sum_{k=0}^n (-1)^k{{n}\choose{k}}x^{n+k-1}dx = \sum_{k=0}^n (-1)^k{{n}\choose{k}}\frac{x^{n+k}}{n+k} $$

Answer 2

An overkill, just for fun. Recalling the Melzak's identity $$f\left(x+y\right)=x\dbinom{x+n}{n}\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{f\left(x+y\right)}{k+x}$$ where $f$ is an algebraic polynomial up to degree $n$ and $x,y\in\mathbb{R},\, x\neq-k,\,k=0,\dots,n$, we can observe, taking $f\equiv1$ and $x=n$, that $$1=n\dbinom{2n}{n}\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{1}{k+n}.$$