Alternating sum of binomial coefficients multiplied by (1/k+1)

Question

I'm trying to prove that

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

So far I've tried induction (which doesn't really work at all), using well known facts such as

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

and trying to apply identities like

$${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}$$

Would anyone be able to point me towards the right method? Should I be looking to apply an identity or is there a method I'm missing?

$$\frac{1}{k+1}\binom{n}{k} = \frac{1}{n+1}\binom{n+1}{k+1}$$ — Daniel Fischer, Oct 08 '14 at 15:24
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 11:07

C.S. · Answer 1 · 2014-10-09T02:19:53.943

9

Note that $$(1-x)^{n} = \sum_{k=0}^{n} {n \choose k}(-1)^{n-k}\ x^{k} =(-1)^{n}\sum_{k=0}^{n}{n\choose k} (-1)^{-k}x^{k} =(-1)^{n}\sum_{k=0}^{n}{n\choose k} (-1)^{k}x^{k}$$ And since $\displaystyle \int_{0}^{1} x^{k} \ dx = \frac{1}{k+1}$. Its worth looking at $\displaystyle \int_{0}^{1} (1-x)^{n} \ dx$

edited Oct 09 '14 at 02:19

answered Oct 08 '14 at 15:23

C.S.

5,528

This is a good hint, but shouldn't the expression for the sum be $$\binom{n}{k}(-1)^{n-k}x^k$$ or $$\binom{n}{k}(-1)^kx^{n-k}$$ right? – cjferes Oct 08 '14 at 15:28
1

@cjferes thanks. I have edited it. – C.S. Oct 08 '14 at 15:40
Note that $\sum_{k=0}^n {n \choose k} (-1)^k x^{k} = (1-x)^{n}$, then integral on $[0,1]$. – Alfred Chern Oct 09 '14 at 02:36
The expression for the sum should be ${n\choose k}(1)^{n-k}\ (-x)^{k}$but not ${n \choose k}(-1)^{n-k}\ x^{k}$ – Alfred Chern Oct 09 '14 at 02:40

score 3 · Answer 2 · answered Oct 08 '14 at 20:09

From $(k+1)\binom{n+1}{k+1} = (n+1)\binom{n}{k}$ we get

\begin{align*} \sum_{k=0}^n \binom{n}{k} (-1)^k \frac{1}{k+1} &= \frac{1}{n+1} \sum_{k=0}^n \binom{n+1}{k+1}(-1)^k \\ &= \frac{1}{n+1} \Bigl( 1 + \sum_{r=0}^{n+1} \binom{n+1}{r} (-1)^{r-1} \Bigr) \\ &= \frac{1}{n+1} \end{align*} as required.

Mark Fischler · Answer 3 · 2014-10-08T16:48:35.070

This is a special case ($x=1$) of the identity $$ \sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k+x} = \frac{1}{x\binom{n+x}{n}} $$ which is proved in Concrete Mathematics, section 5.3, page 188. The outline of the proof is:

Show that the n-th difference of $1/x$ is $$ \frac{(-1)^n}{\binom{n+x}{n}} $$ This is easy using the methods of finite calculus, cf. chapter 2 of that same book; it is analogous to the n-th derivative of $1/x$.

And show that the n-th difference of any function $f(x)$ is given by $$ \Delta^n f(x) = \sum_k \binom{n}{k} (-1)^{n-k} f(x+k) $$

This can be proved by induction, but it also has a cute proof using the concept of the shift operator $E(f(x) = f(x+1)$ and writing $\Delta^n$ as $(E-1)^n$.

Then using $f(x)=1/x$ the result follows.

Masacroso · Answer 4 · 2019-08-11T12:20:36.870

Using generating functions it is straightforward. Note that

$$ \sum_{k\ge 0}\binom{n}k(-1)^k\frac1{k+1}=(-1)^n\sum_{k\ge 0}\binom{n}k(-1)^{n-k}\frac1{k+1}\\ =\left[\frac{x^n}{n!}\right](-1)^n\left(\sum_{k\ge 0}(-1)^k\frac{x^k}{k!}\right)\left(\sum_{k\ge 0}\frac{x^k}{(k+1)!}\right)\\ =\left[\frac{x^n}{n!}\right](-1)^ne^{-x}\cdot\frac{e^x-1}x=\left[\frac{x^n}{n!}\right](-1)^n\frac{1-e^{-x}}x\\ =\left[\frac{x^n}{n!}\right](-1)^n\sum_{k\ge 1}(-1)^{k-1}\frac{x^{k-1}}{k!} =\frac1{n+1} $$

where the notation $[x^n/n!]f(x)$ means "the $n$-th coefficient that appear in the exponential power series expansion of $f$".

Alternating sum of binomial coefficients multiplied by (1/k+1)

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