Find general form for summation expression: $\sum_{k=0}^n {n\choose k}(-1)^k\frac{1}{k+1}$

Question

I am learning about combinatorics and am trying to solve the following problem.

Find the value of

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

for several values of n. What do you think is the value in general? Prove it.

I have carried out some calculations for various values of n and I think that the solution in terms of $n$ is

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

I also think that induction is not a very helpful way to prove this.

I would appreciate any hints (no complete solutions, please!).

Thanks, congratulations and $+1$ for asking only for hints and not for complete solutions. — Claude Leibovici, Oct 23 '15 at 09:56
Several proofs can be found here: How to prove $\sum\limits_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}$? — Martin Sleziak, Jul 26 '17 at 18:14

score 1 · Accepted Answer · answered Oct 23 '15 at 09:33

1

Hint(s): write $\frac{1}{k+1}=\int_{0}^{1}x^k\,dx$, swap the sum and the integral, exploit the binomial theorem, compute the resulting integral through the substitution $x\mapsto 1-x$.

answered Oct 23 '15 at 09:33

Jack D'Aurizio

353,855

Using your hint, I get $\sum_{k=0}^n [{n \choose k}(-1)^k \int_0^1 x^k dx]$. I'm not sure how to take the sum of an integral, though. – Caleb Owusu-Yianoma Oct 23 '15 at 09:54
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@CKKOY: $\sum_{k=0}^{n}\binom{n}{k}(-1)^k x^k = (1-x)^n$. You just have to compute $\int_{0}^{1}(1-x)^n,dx = \int_{0}^{1}x^n,dx.$ – Jack D'Aurizio Oct 23 '15 at 10:13
I've got the answer now! The hints were useful. :) – Caleb Owusu-Yianoma Oct 23 '15 at 11:32

SchrodingersCat · Answer 2 · 2015-10-23T10:20:45.663

1

HINT: Try integrating the following summation

$$(1-x)^n=\sum_{k=0}^n (^n_k)(-1)^kx^k$$ $$\Rightarrow \int^1_0 (1-x)^n dx=\int^1_0 \sum_{k=0}^n (^n_k)(-1)^kx^kdx$$ Put the required value of x and you will get your answer.

edited Oct 23 '15 at 10:20

answered Oct 23 '15 at 09:35

SchrodingersCat

24,472

score 1 · Answer 3 · answered Oct 23 '15 at 15:27

1

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

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

Now Using $$(1+x)^{n+1} = \binom{n+1}{0}+\binom{n+1}{1}x+\binom{n+1}{2}x^2+....+\binom{n+1}{n+1}x^{n+1}$$

answered Oct 23 '15 at 15:27

juantheron

53,015

How do you make the first step (from the LHS of the first equal sign to the RHS of that equal sign), please? – Caleb Owusu-Yianoma Oct 23 '15 at 19:43
Using the formula $\displaystyle \binom{n}{r} = \frac{n}{r}\binom{n-1}{r-1}$ – juantheron Oct 23 '15 at 19:48

Find general form for summation expression: $\sum_{k=0}^n {n\choose k}(-1)^k\frac{1}{k+1}$

3 Answers3