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Prove that $$\binom{n}{1}(1-x)-\frac{1}{2}\binom{n}{2}(1-x)^2+\frac{1}{3}\binom{n}{3}(1-x)^3+\cdots \cdots +(-1)^{n-1}\frac{1}{n}(1-x)^n$$

$$=(1-x)+\frac{1}{2}(1-x^2)+\frac{1}{3}(1-x^3)+\cdots +\frac{1}{n}(1-x^n)$$

what i try

$$\bigg[1-(1-x)\bigg]^n=\binom{n}{0}-\binom{n}{1}(1-x)+\binom{n}{2}(1-x)^2-\cdots +(-1)^n\binom{n}{n}(1-x)^n$$

Integrate with respect to $x$

$$\frac{x^{n+1}}{n+1}-1=-\binom{n}{0}(1-x)+\binom{n}{1}\frac{(1-x)^2}{2}-\binom{n}{2}\frac{(1-x)^3}{3}+\cdots +(-1)^{n-1}\frac{(1-x)^{n+1}}{n+1}$$

How do I solve it? Help me, please!

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    $\begingroup$ Use a direct approach? $(1-x)^k=\sum_{j=0}^k \binom{k}{j}(-x)^j$. Do this for all the terms on the left side and collect all like powers of $x$. You should get right side. $\endgroup$ – herb steinberg Feb 18 at 4:10
  • $\begingroup$ did not ubderstand can you explain in detail way $\endgroup$ – jacky Feb 18 at 15:35
  • $\begingroup$ For example the coefficient of the term$(-x)^j$ is obtained by $T_j=\sum_{k=j}^n\frac{1}{k}\binom{n}{k}\sum_{k=1}^j\binom{k}{j}$, for $j\ne 0$,so we need to show $\sum_{k=1}^n(-1)^{k-1}\frac{1}{k}\binom{n}{k}+\sum_{j=1}^nT_j(-x)^j$ is identical to the right side. $\endgroup$ – herb steinberg Feb 19 at 1:30
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We put

$$S_n(x) = \sum_{p=1}^n \frac{1}{p} {n\choose p} (-1)^{p+1} (1-x)^p.$$

Working first with the coefficient on $[x^q]$ where $1\le q\le n$ we see that it is

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

Now

$${n\choose p} {p\choose q} = \frac{n!}{(n-p)! \times q! \times (p-q)!} = {n\choose q} {n-q\choose n-p}$$

so we find

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

Introducing

$$f(z) = \frac{(-1)^{n-q+1} (n-q)!}{z+q} \prod_{k=0}^{n-q} \frac{1}{z-k}$$

we have

$$\sum_{p=0}^{n-q} \mathrm{Res}_{z=p} f(z) = \sum_{p=0}^{n-q} \frac{(-1)^{n-q+1} (n-q)!}{p+q} \prod_{k=0}^{p-1} \frac{1}{p-k} \prod_{k=p+1}^{n-q} \frac{1}{p-k} \\ = \sum_{p=0}^{n-q} \frac{(-1)^{n-q+1} (n-q)!}{p+q} \frac{1}{p!} (-1)^{n-q-p} \frac{1}{(n-q-p)!} \\ = \sum_{p=0}^{n-q} (-1)^{p+1} \frac{1}{p+q} {n-q\choose p}.$$

This is the target sum omitting the binomial coefficient in front. Now the residue at infinity of $f(z)$ is clearly zero and hence the sum must be (residues sum to zero)

$$- \mathrm{Res}_{z=-q} f(z) = - (-1)^{n-q+1} (n-q)! \prod_{k=0}^{n-q} \frac{1}{-q-k} \\ = - (n-q)! \prod_{k=0}^{n-q} \frac{1}{q+k} = -(n-q)! \frac{(q-1)!}{n!}.$$

Restoring the binomial coefficient in front we thus have

$$[x^q] S_n(x) = -{n\choose q} (n-q)! \frac{(q-1)!}{n!}$$

or alternatively

$$\bbox[5px,border:2px solid #00A000]{ [x^q] S_n(x) = - \frac{1}{q},}$$

as claimed. Comntinuing with the constant coefficient we find

$$[x^0] S_n(x) = \sum_{p=1}^n \frac{1}{p} {n\choose p} (-1)^{p+1} [x^0] (1-x)^p = \sum_{p=1}^n \frac{1}{p} {n\choose p} (-1)^{p+1}.$$

Using the same technique as before we introduce

$$g(z) = \frac{(-1)^{n+1} n!}{z} \prod_{k=0}^{n} \frac{1}{z-k}$$

We get for

$$\sum_{p=1}^n \mathrm{Res}_{z=p} g(z) = \sum_{p=1}^n \frac{(-1)^{n+1} n!}{p} \prod_{k=0}^{p-1} \frac{1}{p-k} \prod_{k=p+1}^{n} \frac{1}{p-k} \\ = \sum_{p=1}^n \frac{(-1)^{n+1} n!}{p} \frac{1}{p!} (-1)^{n-p} \frac{1}{(n-p)!} \\ = \sum_{p=1}^n \frac{1}{p} {n\choose p} (-1)^{p+1}.$$

This is the target sum. Now the residue at infinity is zero so this sum must be equal to

$$- \mathrm{Res}_{z=0} g(z) = - \mathrm{Res}_{z=0} \frac{(-1)^{n+1} n!}{z^2} \prod_{k=1}^{n} \frac{1}{z-k} \\ = (-1)^{n} n! \left.\left( \prod_{k=1}^{n} \frac{1}{z-k}\right)'\right|_{z=0} \\ = (-1)^{n} n! \left.\left( \prod_{k=1}^{n} \frac{1}{z-k} \sum_{k=1}^n \frac{1}{k-z} \right)\right|_{z=0} = (-1)^n n! \times \frac{(-1)^n}{n!} \sum_{k=1}^n \frac{1}{k}.$$

We have shown that (with harmonic numbers)

$$\bbox[5px,border:2px solid #00A000]{ [x^0] S_n(x) = \sum_{k=1}^n \frac{1}{k} = H_n,}$$

which concludes the argument. If desired we may write this as

$$S_n(x) = \sum_{k=1}^n \frac{1}{k} - \sum_{q=1}^n \frac{1}{q} x^q$$

or

$$\bbox[5px,border:2px solid #00A000]{ S_n(x) = \sum_{q=1}^n \frac{1}{q} (1 - x^q).}$$

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