Consider the following binomial summations:
$$\sum_{k=0}^{n}\binom{n}{k}\frac{\left(-1\right)^k }{k+m}\tag{I}$$
Where $m \in \mathbb N_{\ge1}$
$$\sum_{k=1}^{n}\binom{n}{k}\frac{1 }{k+m}\tag{II}$$
Where $m \in \mathbb N_{\ge1}$
$\text{(I)}$
I use the following identity:
$$\binom{n}{k}\frac{1}{k+m}=\frac{1}{m}\binom{n+m}{m}^{-1}\binom{n}{k}\binom{k+m-1}{k}$$
So the sum changes to:
$$\sum_{k=0}^{n}\binom{n}{k}\frac{\left(-1\right)^k }{k+m}=\frac{1}{m}\binom{n+m}{m}^{-1}\sum_{k=0}^{n}\binom{n}{k}\color{red}{\binom{k+m-1}{k}\left(-1\right)^k}$$ $$=\frac{1}{m}\binom{n+m}{m}^{-1}\sum_{k=0}^{n}\binom{n}{n-k}\color{red}{\binom{-m}{k}}$$$$=\frac{1}{m}\binom{n+m}{m}^{-1}\binom{n-m}{n}$$
But is there any closed form when $m=0$?, in other words does there exist any closed form for the following sum:
$$\sum_{k=1}^{n}\binom{n}{k}\frac{\left(-1\right)^k }{k}$$
$\text{(II)}$
$$\sum_{k=1}^{n}\binom{n}{k}\frac{1 }{k+m}=\frac{1}{m}\binom{n+m}{m}^{-1}\sum_{k=1}^{n}\binom{n}{k}\color{red}{\binom{k+m-1}{k}} $$ $$=\frac{1}{m}\binom{n+m}{m}^{-1}\sum_{k=1}^{n}\binom{n}{n-k}\color{red}{\binom{-m}{k}\left(-1\right)^k}$$
The final answer depends on the closed form of alternating sign Vandermonde's convolution, which apparently there does not exist such closed form.
My question is that : what is the closed form of $\text{(II)}$?
For $\text{(II)}$, there is a closed form $$S_n=\sum_{k=1}^{n}\binom{n}{k}\frac{1 }{k+m}=\frac 1m \left(2^n-1+(-1)^{-m} n B_{-1}(m+1,n) \right)\tag 1$$ where appears the incomplete beta function.
For $\text{(I)}$ $$T_n=\sum_{k=1}^{n}\binom{n}{k}\frac{(-1)^k }{k+m}=\frac{\Gamma (m)\, \Gamma (n+1)}{\Gamma (m+n+1)}-\frac{1}{m}\tag 2$$ which, for $m=0$ reduces to $-H_n$.
To make a series expansion of $(2)$ around $m=0$, recall first $$\Gamma(m)=\frac{1}{m}-\gamma +\frac{1}{12} \left(6 \gamma ^2+\pi ^2\right) m+O\left(m^2\right)$$ and $$\Gamma(m+n+1)=\Gamma (n+1)+m \Gamma (n+1) \psi (n+1)+O\left(m^2\right)$$ So $$T_n=-\psi (n+1)-\gamma +O\left(m\right)=-H_n+O\left(m\right)$$