How to prove $\sum\limits_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}$?

Question

Other than the general inductive method,how could we show that $$\sum_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}$$

Apart from induction, I tried with Wolfram Alpha to check the validity, but I can't think of an easy (manual) alternative.

Please suggest an intuitive/easy method.

Answer 1

Look at $$\int_0^1(1-x)^n dx$$ This is easy to compute by substitution.

Now compute it the hard way, by expanding using the Binomial Theorem, and integrating term by term.

Answer 2

Note that $$\begin{align*}\frac{(n+1)\binom{n}{r}}{r+1} &= \frac{(n+1)n!}{(r+1)r!(n-r)!} = \frac{(n+1)!}{(r+1)!(n-r)!}\\ &= \frac{(n+1)!}{(r+1)!((n+1)-(r+1))!} = \binom{n+1}{r+1}.\end{align*}$$ Therefore, $$\begin{align*} (n+1)\sum_{r=0}^{n}(-1)^r\frac{\binom{n}{r}}{r+1} &= \sum_{r=0}^{n}(-1)^r\frac{(n+1)\binom{n}{r}}{r+1}\\ &= \sum_{r=0}^{n}(-1)^r\binom{n+1}{r+1}\\ &= -\sum_{r=0}^n(-1)^{r+1}\binom{n+1}{r+1}\\ &= -\sum_{s=1}^{n+1}(-1)^s\binom{n+1}{s}\qquad(\text{setting }s=r+1)\\ &= \left(-\sum_{s=0}^{n+1}(-1)^s\binom{n+1}{s}\right) + (-1)^0\binom{n+1}{0}\\ &= -(1-1)^{n+1} + 1\\ &= 1. \end{align*}$$

Dividing through by $n+1$ gives the desired result.

Once you realize that $$\frac{(n+1)\binom{n}{r}}{r+1} = \binom{n+1}{r+1},$$ it should be obvious that you are dealing with a binomial expansion of some $(n+1)$st power. Then it's just a matter of figuring out which power, and whether any terms are missing.

Answer 3

Here's a probabilistic proof of the equivalent identity $$\frac{n}{n+1} = \sum\limits_{r=1}^n \frac{(-1)^{r+1}}{r+1}\binom{n}{r} .$$

Choose $n+1$ numbers at random from (the uniform distribution on) $[0,1]$ and call them $x_1, x_2, \ldots, x_{n+1}$. If we select $r$ numbers from $\{x_2, \ldots, x_{n+1}\}$, the probability that $x_1$ is larger than all $r$ of them is $\frac{1}{r+1}$.

Question: What's the probability that at least one of $x_2, \ldots, x_{n+1}$ is larger than $x_1$?

Answer 1: Using the complement, it's $1 - \frac{1}{n+1} = \frac{n}{n+1}$.

Answer 2: By the principle of inclusion-exclusion, it's also $\sum\limits_{r=1}^n (-1)^{r+1}\binom{n}{r} \frac{1}{r+1}.$

(For those not familiar with inclusion-exclusion, it's the generalization of the identity $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ from 2 sets to $n$ sets. First you add in the probabilities of all the singleton sets, but that double-counts the probabilities of the intersections of two sets, so you subtract those off, but then you have to add back the probabilities of the intersections of three sets, etc.)

Answer 4

This an old question, but I recently remembered a nice proof using partial fractions from an exam paper;

Consider the partial fraction expansion $$\frac{n!}{x(x+1)...(x+n)}=\frac{c_{0}}{x} + \frac{c_{1}}{x+1}+...+\frac{c_{n}}{x+n}$$ where $c_{r}$, $r=0,1,2,...n$ are constants. Multiplying both sides by $x+r$ and substituting $x=-r$, we have $$c_{r} = (-1)^{r}\binom{n}{r}$$

Now, sub $x=1$ into the original equation to get

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

Answer 5

My answer use partial integration.