let, n be a positive integer.Then,${C_r}=$$ {n} \choose {r}$.Now evaluate $ {C_0}- {C_1}/2+ {C_2}/3+.....+ (-1)^n {C_n}/(n+1)$
I expand $(1-x)^n$ and integrating both side and putting $x=1$ the required series comes.But it gets 0.But answer is $1/n+1$
One has
$$\int_0^1(1-x)^ndx=\sum_{k=0}^n(-1)^k\binom{n}{k}\int_0^1x^kdx$$
The anti derivative of $(1-x)^n$ is $F(x) =-{1\over n+1}(1-x)^{n+1}$.
The anti derivative of $x^k$ is ${x^{k+1}\over k+1}$.
Substituting $x=1$, one gets
$$\sum_{k=0}^n(-1)^k{\binom{n}{k}\over k+1}=F(1)-F(0)={1\over n+1}$$
Another way:
$$\dfrac{\binom nr}{r+1}=\cdots\dfrac{\binom{n+1}{r+1}}{n+1}$$
$$\sum_{r=0}^n\dfrac{(-1)^r\binom nr}{r+1}=\dfrac1{n+1}\sum_{r=0}^n\binom{n+1}{r+1}$$
Now $$\sum_{r=0}^n(-1)^r\binom{n+1}{r+1}=\binom{n+1}0-(1-1)^{n+1}$$
