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. 
 A: Firstly,
$$\frac{^nC_{1}}{1 \times 2}+\frac{^nC_{2}}{2 \times 3}+\frac{^nC_{3}}{3 \times 4}+..+\frac{^nC_{n}}{n \times (n+1)} =\left[\frac{^nC_{1}}{1}+\frac{^nC_{2}}{2}+\frac{^nC_{3}}{3}+..+\frac{^nC_{n}}{n}\right]-\left[\frac{^nC_{1}}{2}+\frac{^nC_{2}}{3}+\frac{^nC_{3}}{4}+..+\frac{^nC_{n}}{(n+1)}\right] $$
Now consider,
\begin{align*}
(1+x)^n-1 & = \binom{n}{1}x+\binom{n}{2}x^2+\dotsb+\binom{n}{n}x^n\\
\int_0^1((1+x)^n-1)\,dx & = \binom{n}{1}\frac{1}{2}+\binom{n}{2}\frac{1}{3}+\dotsb+\binom{n}{n}\frac{1}{n+1}\\
\frac{(1+x)^n-1}{x} & = \binom{n}{1}1+\binom{n}{2}x+\dotsb+\binom{n}{n}x^{n-1}\\
\int_0^1\frac{(1+x)^n-1}{x}\,dx & = \binom{n}{1}\frac{1}{1}+\binom{n}{2}\frac{1}{2}+\dotsb+\binom{n}{n}\frac{1}{n}.
\end{align*}
So your given series is 
$$\int_0^1\frac{(1+x)^n-1}{x}\,dx-\int_0^1((1+x)^n-1)\,dx$$

Since the question was edited later:

Since you made the changes in the sign of the actual expression, so you need to redo what I have proposed with $(1-x)^n$.
A: 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
$$\frac{1}{n+1}
(- \mathrm{Res}_{z=0} f(z) - \mathrm{Res}_{z=1} f(z) )
= \frac{1}{n+1} (1 + (n+1) H_n - (n+1))
\\ = H_n - \frac{n}{n+1} = H_{n+1} - 1.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\sum_{r = 1}^{n}{\pars{-1}^{r + 1} \over r\pars{r + 1}}
{n \choose r} =
H_{n + 1} - 1:\ {\LARGE ?}}$.

\begin{align}
&\bbox[10px,#ffd]{\sum_{r = 1}^{n}{\pars{-1}^{r + 1} \over r\pars{r + 1}}
{n \choose r}} =
-\sum_{r = 1}^{n}
{n \choose r}\pars{-1}^{r}\pars{\int_{0}^{1}x^{r - 1}\,\dd x}
\pars{\int_{0}^{1}y^{r}\,\dd y}
\\[5mm] = &\
-\int_{0}^{1}\int_{0}^{1}\sum_{r = 1}^{n}{n \choose r}\pars{-xy}^{r}\,
{\dd x \over x}\,\dd y =
-\int_{0}^{1}\int_{0}^{1}\bracks{\pars{1 - xy}^{n} - 1}\,{\dd x \over x}\,\dd y
\\[5mm] = &\
-\int_{0}^{1}\int_{0}^{y}\bracks{\pars{1 - x}^{n} - 1}
\,{\dd x \over x}\,\dd y =
-\int_{0}^{1}{\pars{1 - x}^{n} - 1 \over x}\int_{x}^{1}\dd y\,\dd x
\\[5mm] = &\
-\int_{0}^{1}{\pars{1 - x}^{n + 1} - 1 + x \over x}\,\dd x =
\int_{0}^{1}x\,{x^{n} - 1 \over x - 1}\,\dd x =
\int_{0}^{1}\sum_{k = 1}^{n}x^{k}\,\dd x
\\[5mm] = &\
\sum_{k = 1}^{n}{1 \over k + 1} =
\sum_{k = 2}^{n + 1}{1 \over k} = \bbx{H_{n + 1} - 1} \\ &
\end{align}
