<s>[INST] Compute $(-1)^n\sum_{k=1}^n (-1)^k\frac{(k+n-1)!}{(k-1)!(k-1)!(n-k)!}$ 
*

*Compute $(-1)^n\sum_{k=1}^n (-1)^k\frac{(k+n-1)!}{(k-1)!(k-1)!(n-k)!}$

*Define $a_{k,m}=\frac{(-1)^{k+m}(n+k-1)!(n+m-1)!}{(k+m-1)[(k-1)!(m-1)!]^2(n-m)!(n-k)!}$
Compute $\sum_{k=1}^n\sum_{m=1}^na_{k,m}$
(Hint: one approach is considering the Hilbert matrix. Other approaches are also welcome. Thanks.)
 [/INST] For the first one, suppose we are trying to evaluate
$$(-1)^n \sum_{k=1}^n (-1)^k 
\frac{(k+n-1)!}{(k-1)!(k-1)!(n-k)!}$$
which is
$$(-1)^n \sum_{k=1}^n (-1)^k 
{n\choose k} k {n+k-1\choose k-1}.$$ 
Introduce
$${n+k-1\choose k-1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+k-1}}{z^k} \; dz.$$
This yields for the sum 
$$\frac{(-1)^n}{2\pi i}
\int_{|z|=\epsilon}
(1+z)^{n-1} 
\sum_{k=1}^n (-1)^k {n\choose k} k \frac{(1+z)^k}{z^k}
\; dz.$$
Now observe that
$$x((1+x)^n)' = \sum_{r=1}^n {n\choose r} r x^r
= nx(1+x)^{n-1}.$$
We thus obtain
$$\frac{n\times (-1)^{n+1}}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z}
\left(1-\frac{1+z}{z}\right)^{n-1}
\; dz
\\ = \frac{n\times (-1)^{n+1}}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z}
\frac{(-1)^{n-1}}{z^{n-1}}
\; dz
\\ = \frac{n}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z^n}
\; dz.$$
This evaluates to $$n\times {n\choose n-1} = n^2$$
by inspection.
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