Simplifying $\frac{^nC_{1}}{1 \times 2}-\frac{^nC_{2}}{2 \times 3}+..+(-1)^{n+1}\frac{^nC_{n}}{n \times (n+1)}$ I was thinking about the following sum - 
$\frac{^nC_{1}}{1 \times 2}-\frac{^nC_{2}}{2 \times 3}+\frac{^nC_{3}}{3 \times 4}+..+(-1)^{n+1}\frac{^nC_{n}}{n \times (n+1)} = 1 + \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+..+\frac{1}{n+1}$
How to approach to prove this summation equality?, I could owrite the LHS as $2n - 3n(n-1) + 2n(n-1)(n-2) +...$ after simplification. 

We can prove the following involving harmonic numbers:
$$\sum_{r=1}^n \frac{(-1)^{r+1}}{r(r+1)} {n\choose r}
= H_{n+1} - 1.$$
Start by writing for the LHS
$$\sum_{r=1}^n \frac{(-1)^{r+1}}{r(r+1)} \frac{r+1}{n+1} {n+1\choose r+1}
= \frac{1}{n+1}
\sum_{r=1}^n \frac{(-1)^{r+1}}{r} {n+1\choose r+1}
\\ = \frac{1}{n+1}
\sum_{r=2}^{n+1} \frac{(-1)^r}{r-1} {n+1\choose r}.$$
Working with the sum term and introducing
$$f(z) =  (-1)^{n+1}
\frac{(n+1)!}{z-1} \prod_{q=0}^{n+1} \frac{1}{z-q}$$
we have for $2\le r\le n+1$ that
$$\sum_{r=2}^{n+1} \mathrm{Res}_{z=r} f(z)
= \sum_{r=2}^{n+1}  (-1)^{n+1}  \frac{(n+1)!}{r-1}
\prod_{q=0}^{r-1} \frac{1}{r-q} \prod_{q=r+1}^{n+1} \frac{1}{r-q}
\\ = \sum_{r=2}^{n+1}  (-1)^{n+1}  \frac{(n+1)!}{r-1}
\frac{1}{r!} \frac{(-1)^{n-r+1}}{(n+1-r)!}
= \sum_{r=2}^{n+1} \frac{(-1)^{r}}{r-1} {n+1\choose r},$$
so this is the  desired sum. Now residues sum to  zero and the residue
at infinity  is zero since $\lim_{R\to\infty}  2\pi R / R  / R^{n+2} =
0.$ The remaining poles are at $z=0$ and $z=1.$ We write
$$f(z) =  (-1)^{n+1} (n+1)! \frac{1}{z} \frac{1}{(z-1)^2}
 \prod_{q=2}^{n+1} \frac{1}{z-q}$$
to get
$$\mathrm{Res}_{z=0} f(z)
= (-1)^{n+1} (n+1)! \frac{(-1)^n}{(n+1)!} = -1$$
as well as
$$\mathrm{Res}_{z=1} f(z)
= (-1)^{n+1} (n+1)!
\left. \left(-\frac{1}{z^2} \prod_{q=2}^{n+1} \frac{1}{z-q}
- \frac{1}{z} \prod_{q=2}^{n+1} \frac{1}{z-q}
\sum_{q=2}^{n+1} \frac{1}{z-q} \right) \right|_{z=1}
\\ = (-1)^{n+1} (n+1)!
\left(- \frac{(-1)^n}{n!}
+ \frac{(-1)^n}{n!} H_n \right)
= n+1 - (n+1) H_n.$$
Restoring the multiplier in front we thus obtain
$$\frac{1}{n+1}
(- \mathrm{Res}_{z=0} f(z) - \mathrm{Res}_{z=1} f(z) )
= \frac{1}{n+1} (1 + (n+1) H_n - (n+1))
\\ = H_n - \frac{n}{n+1} = H_{n+1} - 1.$$
