Evaluate the sum ${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}$ Evaluate the sum $${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}.$$
I have tried comparing this to the similar problem here.
I believe I need to differentiate or integrate? But I'm not sure how that might work.
Any ideas? Thanks.
 A: This is $$\sum_{k=0}^{n}\frac{(-1)^{k}}{k+1}\binom{n}{k}.$$ Then multiplying each term by $\frac{n+1}{n+1},$ we get $$\sum_{k=0}^{n}\frac{(-1)^{k}}{n+1}\binom{n+1}{k+1}=\frac{1}{n+1}\sum_{k=1}^{n+1}(-1)^{k-1}\binom{n+1}{k}.$$ Adding and subtracting $1$ and applying the Binomial Theorem gives $$\frac{1}{n+1}-\frac{1}{n+1}\sum_{k=0}^{n+1}\binom{n+1}{k}(-1)^{k}1^{n+1-k}=\frac{1}{n+1}(1-(1-1)^{n+1})=\frac{1}{n+1}.$$
A: $$\begin{align}
S&=1-\frac 12\binom n1+\frac 13 \binom n2-\cdots+(-1)^n\frac 1{n+1}\binom nn\\
\times (n+1):\hspace{1cm}\\
(n+1)S&=(n+1)-\frac {n+1}2\binom n1+\frac {n+1}3\binom n2-\cdots+(-1)^n\frac {n+1}{n+1}\binom nn\\
&=\binom {n+1}1-\binom {n+1}2+\binom {n+1}3-\cdots +(-1)^n\binom {n+1}{n+1}\\
&=\color{blue}{\binom {n+1}0}\underbrace{\color{blue}{-\binom {n+1}0}+\binom {n+1}1-\binom {n+1}2+\binom {n+1}3-\cdots +(-1)^n\binom {n+1}{n+1}}_{=-\sum_{r=0}^{n+1}\binom {n+1}r(-1)^r=-(1-1)^{n+1}=0}\\
&=1\\
S&=\color{red}{\frac 1{n+1}}
\end{align}$$
A: Hint
$${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}.=\sum_{j=0}^{n} \frac{(-1)^j}{j+1} \binom n j $$
$$\sum_{j=0}^{n} \frac{(-1)^j}{j+1} x^{j+1} \binom n j =\int \sum_{j=0}^{n} (-1)^j x^{j}\binom n j =\int (1-x)^n=- \frac { (1-x)^{n+1}} {n+1} $$
$$\sum_{j=0}^{n} \frac{(-1)^j}{j+1} \binom n j =\int_0^1 (1-x)^n=\frac 1 {n+1}$$
