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. 

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 $(1-x)^n$.
