wang yin <shredder@sohu.com> writes:
> If I express "k to the m falling" as k!/(k-m)! in nusum, I
> get:
>
> (C1) nusum(k!/(k-m)!,k,0,n);
>
> (n + 1)! m
> (D1) ---------------- + --------------
> (m + 1) (n - m)! (m + 1) (- m)!
>
> How can I abtain the desired result
>
> "n to the (m+1) falling"/(m+1) ?
But Maxima's result is correct. I assume that you actually want to
calculate
(C1) nusum(k!/(k-m)!,k,m,n);
(n + 1)!
(D1) ----------------
(m + 1) (n - m)!
for n>=m>=0. This result is (n+1 to the m+1 falling)/(m+1), which is
easily proven to be correct by induction wrt. n for each fixed m (you
might prefer to introduce first a new summation index l=k-m instead).
Note that the result you got is also correct since 1/(- m)! =
(1/gamma)(1-m) = 0 for an integer m>=1. Your original sum also
contains such terms in general.
Wolfgang