Sum of positive terms gives negative answer

Bug introduced in 7.0 and fixed in 9.0

Mathematica evaluates Sum[((n - y - 1)*(n - y)^2*n^y)/y!, {y, 0, n - 2}] as -2 e^n n. This should not be a negative value. What am I doing wrong?

• This is a problem of version 8. In version 9.04 I get (1/(n! Gamma[ n]))(-n! (5 n^n + E^n (1 + 4 n) Gamma[n] - E^n Gamma[n, n] - 2 E^n Gamma[1 + n, n]) + n^n Gamma[n] HypergeometricPFQ[{2, 2, 2}, {1, 1, 1 + n}, n]). – Sjoerd C. de Vries Mar 11 '13 at 22:34
• OK. HypergeometricPFQ[{2, 2, 2}, {1, 1, 1 + n}, n] looks very hard to interpret! Do you have any idea what that looks like asymptotically? – Lembik Mar 11 '13 at 22:38
• @Sjoerd Interestingly, if you simply change the upper limit from n-2 to n-1 (which introduces a zero into the sum, leaving it unaltered) MMA 8 obtains (1/((1 + n)! Gamma[ 2 + n]))n ((1 + n)! (n^n (1 + 2 n) - E^n Gamma[2 + n] - E^n Gamma[2 + n, n]) + n^n Gamma[2 + n] HypergeometricPFQ[{2, 2, 2}, {1, 1, 2 + n}, n])--which numerical tables show is correct. Also, if the full Sum is involved within a more complex expression, it might get evaluated correctly. – whuber Mar 11 '13 at 22:39
• Lembik, hypergeometric functions are nice: they have series expansions, known poles, integral representations, etc. Take a look at the help page for the series definition. – whuber Mar 11 '13 at 22:40
• @whuber Fact remains that this particular result is indeed dead wrong. – Sjoerd C. de Vries Mar 11 '13 at 22:40

This is apparently a problem of version 8 where I get the same. In version 9.0.1 I get

(1/(n! Gamma[ n]))(-n! (5 n^n + E^n (1 + 4 n) Gamma[n] - E^n Gamma[n, n] -
2 E^n Gamma[1 + n, n]) + n^n Gamma[n] HypergeometricPFQ[{2, 2, 2}, {1, 1, 1 + n}, n])

or, with nicer formatting, This can be reduced with

Sum[((n - y - 1)*(n - y)^2*n^y)/y!, {y, 0, n - 2}] // FullSimplify

to

(2 n (n^n - E^n Gamma[n, n]))/Gamma[n] 