I need to show that
$$\sum_{n=0}^x(-1)^n {x\choose n}\left(\frac{1}{n+l-1}-\frac{1}{n+l}\right) >0,$$
for any $l>0$. I tried to prove this, but I didn't get anywhere.
Apparently the above sum is equal to $\frac{(l-2)!(x+1)!}{(x+l)!}$. However I have no idea how to prove this either. Any tips on how to approach this would be greatly appreciated!
Since $\frac{1}{a}=\int_{0}^{1} z^{a-1}\,dz $, we have:
$$\begin{eqnarray*} \sum_{n=0}^{x}\binom{x}{n}(-1)^n\left(\frac{1}{n+l-1}-\frac{1}{n+l}\right) &=& \int_{0}^{1}\sum_{n=0}^{x}\binom{x}{n}(-1)^n\left(z^{n+l-2}-z^{n+l-1}\right) \,dz\\&=&\int_{0}^{1}(1-z)^x\left(z^{l-2}-z^{l-1}\right)\,dz\\&=&\int_{0}^{1}(1-z)^{x+1} z^{l-2}\,dz.\end{eqnarray*} $$
The last integral is clearly positive and can be computed through Euler's beta function, or simply by integration by parts.