Harmonic series and the binomial theorem I recently came across the following identity relating to the harmonic series: $\displaystyle \sum_{r=1}^{n}\frac{(-1)^{r+1}C_r^{n}}{r} =\sum_{r=1}^{n}\frac{1}{r}$. It was the first time I had seen this identity, and it got me thinking: are there any more interesting results relating to the sum $\displaystyle \sum_{r=1}^{n}\frac{(-1)^{r+1}C_r^{n}}{r}$ that we can deduce from this relationship? One that is pretty immediate is the fact that this sum must diverge, but I'd be interested to see if anyone could come up with some less-obvious or creative ones.  
 A: Remark : the exact identity is $\sum_{r=1}^n \frac{(-1)^{\mathbf{r-1}}\binom nr}{r} = \sum_{r=1}^n \frac 1r$.
One way to prove this identity is induction.
Another way is as follows :

*

*start with $f:x\mapsto \sum_{r=1}^n \binom nr\frac{(-1)^{r-1}x^r}{r}$;

*$f$ being a polynomial function is differentiable with $f'(x)=\sum_{r=1}^n \binom nr x^{r-1} = \frac1x \left[1-(1-x)^n\right]$ (prolongated with $f'(0)=n$).

*so $f(x)=f(0)+\int_0^x f'(t){\rm d}t$. Let's do that computation :
\begin{align}f(x)&=\int_0^x \frac{1-(1-t)^n}{t}{\rm d}t \\ 
      &\underset{u=1-t}{=} -\int_1^{1-x} \frac{1-u^n}{1-u}{\rm d}u \\ 
      &= -\int_1^{1-x}(1+u+u^2+\dots+u^{n-1}){\rm d}u \\
      &= -\left[u+\frac{u^2}{2}+\dots+\frac{u^n}{n}\right]_1^{1-x} \\
      &= (1+\frac12+\dots+\frac1n)-((1-x)+\frac{(1-x)^2}{2}+\dots + 
         \frac{(1-x)^n}{n})
   \end{align}

*the value $f(1)$ gives the identity.

I think this technique can be generalized, and that it could prove a more general formula obtained by replacing $\frac{x^r}{r}$ by $\frac{x^{r+\alpha}}{r+\alpha}$ in the definition of function $f$...
A: For integer $n$ and real  $x \not\in \{-n, -n+1, \ldots, -1, 0\}$
$$
\sum_r \frac{ (-1)^r  }{x+r}\binom{n}{r} = \frac{1}{x} \frac{1}{\binom{n+x}{n}}
$$
where the sum over $r$ will be a finite sum if $n$ is an integer.  This looks like your identity, but includes the $r=0$ term which yours, of course, does not.
