conversion of Binomial identity into series sum 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!
 A: 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
$$\bbox[5px,border:2px solid #00A000]{
S_n(x) = \sum_{q=1}^n \frac{1}{q} (1 - x^q).}$$
