Hi:

I think I've got it pretty well figured out now - Maple is giving the wrong answer. See below:

The summand "C" is a particular combination of harmonic numbers. When Maple is asked to numerically sum from 0 to infinity it returns test0 = 0. Summing from 1 to infinity (which should give the same result because the k=0 term vanishes), returns test1=unevaluated - Maple seems to do that when a series converges very slowly, and it's not sure.

If you sum from zero to infinity without simplification, it returns unevaluated, but simplification by itself does not return zero, so I theorize that the problem is in the combination of "evalf(simplify(...)) when the first term vanishes.

I have also checked with varying upper limits to 10,000 and the answer is converging to something like -.12, which is reasonable.

I have reported this to technical support.

Too bad - the result could have been zero and I thought Maple had some great new way to determine such things.

____________________________________________________________________

C := -(-H(k+1/2)*H(k+1)+H(1,k+1/2)+H(k+1/2)^2)*GAMMA(k+1/2)/(2*k+1)/Pi^(1/2)/GAMMA(k+1);

> C0:=combine(subs(H(k+1/2)=Psi(k+1/2)-Psi(1/2),H(1,k+1/2)=Psi(1,k+1/2)-Psi(1,1/2),H(k+1)=Psi(k+1)-Psi(1),k=0,C));

> C1:=combine(subs(H(k+1/2)=Psi(k+1/2)-Psi(1/2),H(1,k+1/2)=Psi(1,k+1/2)-Psi(1,1/2),H(k+1)=Psi(k+1)-Psi(1),k=1,C));

2

C := - (-H(k + 1/2) H(k + 1) + H(1, k + 1/2) + H(k + 1/2) )

/ 1/2

GAMMA(k + 1/2) / ((2 k + 1) Pi GAMMA(k + 1))

/

C0 := 0

C1 := 1/3

> C1a := Sum(C,k = 0 .. infinity);

> C1b := Sum(C,k = 1 .. infinity);

infinity

-----

\ /

C1a := ) |-

/ \

-----

k = 0

2

(-H(k + 1/2) H(k + 1) + H(1, k + 1/2) + H(k + 1/2) )

/ 1/2 \

GAMMA(k + 1/2) / ((2 k + 1) Pi GAMMA(k + 1))|

/ /

infinity

-----

\ /

C1b := ) |-

/ \

-----

k = 1

2

(-H(k + 1/2) H(k + 1) + H(1, k + 1/2) + H(k + 1/2) )

/ 1/2 \

GAMMA(k + 1/2) / ((2 k + 1) Pi GAMMA(k + 1))|

/ /

> test0:=evalf(simplify(subs(H(k+1/2)=Psi(k+1/2)-Psi(1/2),H(1,k+1/2)=Psi(1,k+1/2)-Psi(1,1/2),H(k+1)=Psi(k+1)-Psi(1),C1a)));

> test1:=evalf(simplify(subs(H(k+1/2)=Psi(k+1/2)-Psi(1/2),H(1,k+1/2)=Psi(1,k+1/2)-Psi(1,1/2),H(k+1)=Psi(k+1)-Psi(1),C1b)));

test0 := 0.

infinity

-----

\ /

test1 := ) |-1/2 (-2 Psi(k + 1/2) Psi(k + 1)

/ \

-----

k = 1

+ 2 Psi(k + 1/2) gamma - 2 gamma Psi(k + 1)

- 4 ln(2) Psi(k + 1) + 4 ln(2) gamma + 2 Psi(1, k + 1/2)

2 2 2

- Pi + 2 Psi(k + 1/2) + 8 Psi(k + 1/2) ln(2) + 8 ln(2) )

/ 1/2 \

GAMMA(k + 1/2) / ((2 k + 1) Pi GAMMA(k + 1))|

/ /

> testx1:=evalf((subs(H(k+1/2)=Psi(k+1/2)-Psi(1/2),H(1,k+1/2)=Psi(1,k+1/2)-Psi(1,1/2),H(k+1)=Psi(k+1)-Psi(1),C1a)));

> testx2:=(simplify(subs(H(k+1/2)=Psi(k+1/2)-Psi(1/2),H(1,k+1/2)=Psi(1,k+1/2)-Psi(1,1/2),H(k+1)=Psi(k+1)-Psi(1),C1a)));

infinity

----- / /

\ | |

testx1 := ) |- |

/ \ \

-----

k = 0

-(Psi(k + 1/2) + gamma + 2 ln(2)) (Psi(k + 1) + gamma)

2 \

Pi 2|

+ Psi(1, k + 1/2) - --- + (Psi(k + 1/2) + gamma + 2 ln(2)) |

2 /

\

/ 1/2 |

GAMMA(k + 1/2) / ((2 k + 1) Pi GAMMA(k + 1))|

/ /

infinity

-----

\ /

testx2 := ) |-1/2 (-2 Psi(k + 1/2) Psi(k + 1)

/ \

-----

k = 0

+ 2 Psi(k + 1/2) gamma - 2 gamma Psi(k + 1)

- 4 ln(2) Psi(k + 1) + 4 ln(2) gamma + 2 Psi(1, k + 1/2)

2 2 2

- Pi + 2 Psi(k + 1/2) + 8 Psi(k + 1/2) ln(2) + 8 ln(2) )

/ 1/2 \

GAMMA(k + 1/2) / ((2 k + 1) Pi GAMMA(k + 1))|

/ /

>