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An exercise I found somewhere (unfortunately I don't remember where) asks to prove the equality $$\sum_{k=1}^{n-1} \frac{(-1)^{k-1}}{k} \binom{n}{k}\left(1 - \frac{k}{n}\right)^n=\sum_{k=2}^n {1\over k} \, .$$ Below is a description of my attempt to prove it

I tried to evaluate the sum by applying first the Newton formula to $(y-x)^n$, removing the extreme terms of the sum, then dividing by $x$ and integrating over $x\in[0,1]$ and I get an expression of $\sum_{k=1}^n \frac{(-1)^{k}}{k} \binom{n}{k}y^{n-k}$.

The next step is to differentiate over $y$ $n$ times (and by multplying by $y$ at each step) in order to get the $(n-k)^n$ in the sum, but the computations seem very tedious, unless I missed something.

Maybe the method is not the good one? Thanks for your help if you have any clue!

Martin R
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dionysos
  • 695

2 Answers2

1

We seek to evaluate

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

Expanding the powered term,

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

Now for $q=0$ we get

$$\sum_{k=1}^{n-1} \frac{(-1)^{k+1}}{k} {n\choose k} = \sum_{k=1}^{n-1} \frac{(-1)^{n-k+1}}{n-k} {n\choose k} \\ = [z^n] \log\frac{1}{1-z} \sum_{k=1}^{n-1} (-1)^{n-k+1} z^k {n\choose k}.$$

Here we may include $k=n$ because it does not get past the extractor. We get $(-1)^{n+1}/n$ from the $k=0$ term,

$$(-1)^n/n+ (-1)^{n+1} [z^n] \log\frac{1}{1-z} (1-z)^n \\ = (-1)^n/n - [z^n] \log\frac{1}{1-z} (z-1)^n = (-1)^n/n - \;\underset{z}{\mathrm{res}}\; \frac{1}{z^{n+1}} (z-1)^n \log\frac{1}{1-z}.$$

Now put $z/(z-1)=u$ in the residue so that $z=u/(u-1)$ and $dz = -1/(u-1)^2 \; du$ to get (flip sign)

$$\;\underset{u}{\mathrm{res}}\; \frac{1}{u^{n+1}} (u-1) \log\frac{1}{1-u/(u-1)} \frac{1}{(u-1)^2} \\ = \;\underset{u}{\mathrm{res}}\; \frac{1}{u^{n+1}} \log(1-u) \frac{1}{u-1} \\ = \;\underset{u}{\mathrm{res}}\; \frac{1}{u^{n+1}} \log\frac{1}{1-u} \frac{1}{1-u} = H_n.$$

We thus have

$$\frac{(-1)^n}{n} + H_n.$$

The remaining sum is

$$\sum_{q=1}^n {n\choose q} (-1)^q \frac{1}{n^q} \sum_{k=1}^{n-1} (-1)^{k+1} {n\choose k} k^{q-1} \\ = \sum_{q=1}^{n} {n\choose q} (-1)^{q} \frac{1}{n^q} (q-1)! [z^{q-1}] \sum_{k=1}^{n-1} (-1)^{k+1} {n\choose k} \exp(kz) \\ = - \sum_{q=1}^{n} {n\choose q} (-1)^{q} \frac{1}{n^q} (q-1)! [z^{q-1}] \left[(1-\exp(z))^n - 1 - (-1)^n \exp(nz) \right] \\ = -1 + \frac{(-1)^n}{n} \sum_{q=1}^{n} {n\choose q} (-1)^q + (-1)^n n! \sum_{q=1}^{n} {n\choose q} (-1)^q \frac{1}{n^q} {q-1\brace n}.$$

Now with the third term we require $n\ge q$ as well as $q-1\ge n$ which makes for a zero contribution. Collecting everything we find

$$\frac{(-1)^n}{n} + H_n - 1 + \frac{(-1)^n}{n} ((1-1)^n-1) = H_n-1.$$

This is the claim. We could have skipped the Stirling numbers just noting that $(1-\exp(z))^n = (-1)^n z^n +\cdots$ so we get zero from the coefficient $[z^{q-1}] (1-\exp(z))^n$ because $0\le q-1\le n-1.$

Marko Riedel
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0

One can combine the results from

  1. Prove $\sum_{k=1}^n \frac{(-1)^{k-1}}{k }\binom{n}{k}= H_n$ without the Beta function and
  2. A property of every real polynomial.

For fixed $n$ is $$ \left( 1-\frac kn\right)^n = 1 - k P(k) $$ where $P$ is a polynomial of degree $n-1$ with $P(0) = 1$. It follows that $$ \sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} \binom{n}{k}\left(1 - \frac{k}{n}\right)^n = \sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} \binom{n}{k} + \sum_{k=1}^{n}(-1)^{k} \binom{n}{k} P(k) \\ = -P(0) + \sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} \binom{n}{k} + \sum_{k=0}^{n}(-1)^{k} \binom{n}{k} P(k) \, . $$ The first sum on the right is equal to the $n$-th harmonic number $H_n= \sum_{k=1}^n 1/k$ as shown in the first linked reference, and the second sum is zero as shown in the second reference (it is the $n$-th iterated forward difference of a polynomial of degree less than $n$). It follows that $$ \sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} \binom{n}{k}\left(1 - \frac{k}{n}\right)^n = -1 + H_n = \sum_{k=2}^n \frac 1k $$ which is the desired identity. (Note that including the summand for $k=n$ does not change the sum on the left.)

Martin R
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