TAOCP 5.1.3 Exercise 6
The mistake is that Mr.
Exercise 6. [M21] [M21] Mr. B. C. Dull noticed that, by Eqs. (4) and (13),
$$ n! = \sum_{k>0} (-1)^k \sum_{j>0} \binom{j}{k} \binom{n+j}{n}. $$
Carrying out the sum on $k$ first, he found that
$$ \sum_{k \ge 0} (-1)^k \binom{j}{k} = 0 $$
for all $j > 0$; hence $n! = 0$ for all $n > 0$. Did he make a mistake?
Verified: yes
Solve time: 36m38s
The mistake is that Mr. Dull replaced the inner sum
$$ \sum_{k>0} (-1)^k \binom{j}{k} $$
by
$$ \sum_{k\ge0} (-1)^k \binom{j}{k}, $$
thereby introducing the $k=0$ term. That term is essential.
Starting from
$$ n! = \sum_{k>0} (-1)^k \sum_{j>0} \binom{j}{k}\binom{n+j}{n}, $$
suppose we interchange the order of summation. Since $\binom{j}{k}=0$ when $k>j$, for each fixed $j$ only finitely many values of $k$ contribute, so the inner sum is
$$ \sum_{k>0} (-1)^k \binom{j}{k}. $$
Now
$$ \sum_{k\ge0} (-1)^k \binom{j}{k} =(1-1)^j=0 \qquad (j>0), $$
hence
$$ \sum_{k>0} (-1)^k \binom{j}{k}
-,\binom{j}{0} =-1. $$
Indeed,
$$ 0
\binom{j}{0} +\sum_{k>0} (-1)^k \binom{j}{k}
1+\sum_{k>0} (-1)^k \binom{j}{k}, $$
so the latter sum equals $-1$.
Therefore
$$ n!
\sum_{j>0} \left(\sum_{k>0} (-1)^k \binom{j}{k}\right) \binom{n+j}{n}
-\sum_{j>0}\binom{n+j}{n}. $$
At this point one immediately sees another reason Mr. Dull's argument cannot be valid: the resulting series is divergent. The original double series is not absolutely convergent, so rearranging and evaluating it by treating the rows independently is not justified. The identities from Eqs. (4) and (13) arise from formal manipulations of infinite series, and changing the order of summation changes the value.
Thus Mr. Dull made a mistake. The identity
$$ \sum_{k\ge0} (-1)^k \binom{j}{k}=0 $$
is correct, but it does not apply to the sum occurring in the formula, which is over $k>0$. The omitted $k=0$ term contributes $1$, giving
$$ \sum_{k>0} (-1)^k \binom{j}{k}=-1, $$
not $0$. Moreover, the subsequent interchange of infinite summations is not legitimate. Hence the conclusion $n!=0$ is false.