Evaluate the sum ${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}$

Question

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.

$n$ is a variable here. I don't need a numeric value, but rather a closed form (not a sum). — Tyler Durden, Oct 10 '17 at 03:09
@TylerDurden I used the search box with the terms alternating binomial coefficient. The link is on the second page of results. — grand_chat, Oct 10 '17 at 05:19
See also How to prove $\sum\limits_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}$? and other posts linked there. — Martin Sleziak, Oct 10 '17 at 11:08

score 8 · Accepted Answer · answered Oct 10 '17 at 03:20

8

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 $$\frac{1}{n+1}-\frac{1}{n+1}\sum_{k=0}^{n+1}\binom{n+1}{k}(-1)^{k}1^{n+1-k}=\frac{1}{n+1}(1-(1-1)^{n+1})=\frac{1}{n+1}.$$

answered Oct 10 '17 at 03:20

RideTheWavelet

3,543

Thanks so much for your answer. When you say "Adding and subtracting $1$," could you clarify what that means? – Tyler Durden Oct 10 '17 at 03:32
1

Yep: We have a $k=0$ term missing in the previous line, and we observe that this is just $(-1)^{0-1}\binom{n+1}{0}=-1,$ so we add $1$ and subtract $1$ (in the form of the $k=0$ term), and of course, scale both by the multiplicative factor $1/(n+1)$ which is multiplying the rest of the sum. – RideTheWavelet Oct 10 '17 at 03:36
1

Since we've really just added $0,$ the sum doesn't change, but having the $k=0$ term allows us to apply the Binomial Theorem. Also, love the user name, but let's not talk about that. And not talk about it. – RideTheWavelet Oct 10 '17 at 03:39
Since $\frac{1}{n+1}(\sum_{k=1}^{n+1} (-1)^{k-1} {n+1 \choose k}) = \frac{1}{n+1}(1 + \sum_{k=0}^{n+1} (-1)^{k-1} {n+1 \choose k})$, where are you getting that negative sign in $\frac{1}{n+1} - \frac{1}{n+1} \ldots$ from? – Tyler Durden Oct 10 '17 at 03:42
Ah, you multiply the sum by $-1$ and add $1$ to the coefficient on $-1$ in the sum to make it $(-1)^k$. I see now. – Tyler Durden Oct 10 '17 at 03:44
Yes, that's exactly right. – RideTheWavelet Oct 10 '17 at 04:30

score 1 · Answer 2 · answered Oct 10 '17 at 05:40

$$\begin{align} S&=1-\frac 12\binom n1+\frac 13 \binom n2-\cdots+(-1)^n\frac 1{n+1}\binom nn\\ \times (n+1):\hspace{1cm}\\ (n+1)S&=(n+1)-\frac {n+1}2\binom n1+\frac {n+1}3\binom n2-\cdots+(-1)^n\frac {n+1}{n+1}\binom nn\\ &=\binom {n+1}1-\binom {n+1}2+\binom {n+1}3-\cdots +(-1)^n\binom {n+1}{n+1}\\ &=\color{blue}{\binom {n+1}0}\underbrace{\color{blue}{-\binom {n+1}0}+\binom {n+1}1-\binom {n+1}2+\binom {n+1}3-\cdots +(-1)^n\binom {n+1}{n+1}}_{=-\sum_{r=0}^{n+1}\binom {n+1}r(-1)^r=-(1-1)^{n+1}=0}\\ &=1\\ S&=\color{red}{\frac 1{n+1}} \end{align}$$

user577215664 · Answer 3 · 2017-10-10T04:01:51.940

0

Hint

$${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}.=\sum_{j=0}^{n} \frac{(-1)^j}{j+1} \binom n j $$

$$\sum_{j=0}^{n} \frac{(-1)^j}{j+1} x^{j+1} \binom n j =\int \sum_{j=0}^{n} (-1)^j x^{j}\binom n j =\int (1-x)^n=- \frac { (1-x)^{n+1}} {n+1} $$

$$\sum_{j=0}^{n} \frac{(-1)^j}{j+1} \binom n j =\int_0^1 (1-x)^n=\frac 1 {n+1}$$

edited Oct 10 '17 at 04:01

answered Oct 10 '17 at 03:31

user577215664

40,625

Evaluate the sum ${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}$

3 Answers3