Algebraic identity arising from coupon collectors problem.

Question

Prove the following identity:

$$\sum\limits_{k=1}^{n} (-1)^{k+1}\frac{n \choose k}{k}= 1+\frac 1 2 + \frac 1 3 + \dots \frac 1 n \tag{1}$$

EDIT: Equation (1) above is proved via integration by @donaldsplutterwit. And I suppose this could lead to the equivalent identity below. However, still want to see if there is a more direct way to prove it.

Or equivalently, prove that when $n$ is even (see my attempt for why equivalent):

$$\sum\limits_{k=1}^{n-1}-1^{k+1}\frac{n-1 \choose k}{n-k} = \frac 2 n$$

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My attempt:

I denote $S_n$ as the LHS of equation (1) and try to show $S_n-S_{n-1}=\frac 1 n$

Let

$$S_n = \sum\limits_{k=1}^{n}(-1)^{k+1}\frac{{n \choose k}}{k} = \sum\limits_{k=1}^{n-1}(-1)^{k+1}\frac{{n \choose k}}{k}+\frac{-1^{n+1}}{n}\tag{2}$$

Now,

$${n \choose k} = \frac{n}{n-k} {n-1 \choose k} = \left(1+\frac{k}{n-k}\right){n-1\choose k}$$

Substituting into equation (2):

$$S_n = S_{n-1} + \sum\limits_{k=1}^{n-1}(-1)^{k+1}\frac{{n-1 \choose k}}{n-k}+\frac{-1^{n+1}}{n}$$

We require: $$S_n-S_{n-1} = \sum\limits_{k=1}^{n-1}(-1)^{k+1}\frac{{n-1 \choose k}}{n-k}+\frac{-1^{n+1}}{n}=\frac1 n$$

This would imply:

$$\sum\limits_{k=1}^{n-1}(-1)^{k+1}\frac{{n-1 \choose k}}{n-k} = \begin{cases}\frac2 n, & n\%2=0\\ 0, & n\%2=1\end{cases}$$

The second case is easy to see since if $n$ is odd, the $k$ th term of the summation above becomes:

$$t_k = -1^{k+1}\frac{n-1 \choose k}{n-k}=-1^{k+1}\frac{(n-1)!}{k!(n-k)!}$$

And

$$t_{n-k} = -1^{n-k+1}\frac{n-1 \choose n-k}{k} = -1^k \frac{(n-1)!}{(n-k)!k!}=-t_k$$

So, every $t_k$ term is matched with a $t_{n-k}$ terms and they all cancel. Haven't managed to make progress on the case when $n$ is even.

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Where did this identity come from? I was studying the Coupon collector's problem and for the case when all coupons have equal probabilities of being collected, we get the average number of draws until all coupons are collected:

$$E(X) = n \sum_i \frac 1 i$$

If the coupons have different probabilities given by $p_i$ we get:

$$E(X) = \sum_i \frac 1 p_i - \sum_{i<j}\frac 1 {p_i+p_j}+\dots +(-1)^{n+1}\frac{1}{p_1+\dots +p_n}$$

Substituting $p_i=\frac 1 n$ for all $i$ into the second equation and substituting into the first leads to equation (1).

Answer 1

Use \begin{eqnarray*} \frac{1}{k}=\int_0^1 x^{k-1} dx. \end{eqnarray*} The sum becomes \begin{eqnarray*} \sum_{k=1}^{n} \frac{(-1)^{k+1}}{k} \binom{n}{k} &=& \sum_{k=1}^{n} \binom{n}{k} (-1)^{k+1} \int_0^1 x^{k-1} dx\\ &=& \int_0^1 \frac{1-(1-x)^{n}}{x}dx. \\ \end{eqnarray*} Now substitute $u=1-x$.

Edit: A similar trick can be used to show the second identity \begin{eqnarray*} \frac{1}{n-k}=\int_0^1 x^{n-k-1}dx. \end{eqnarray*} The sum becomes \begin{eqnarray*} \sum_{k=1}^{n-1} \frac{(-1)^{k+1}}{n-k} \binom{n-1}{k} &=& \sum_{k=1}^{n-1} \binom{n-1}{k} (-1)^{k+1} \int_0^1 x^{n-k-1}dx \\ &=& (-1)^n \int_0^1 \left( (1-x)^{n-1} +(-1)^n x^{n-1}\right) dx. \\ \end{eqnarray*}