Re: An exact 1-D summation challenge - 3

From: G. A. Edgar (edgar_at_math.ohio-state.edu.invalid)
Date: 03/11/05 

Date: Fri, 11 Mar 2005 10:08:37 -0500

>
> > limit(hypergeom([1,2,3,5/2+z], [5/2,2+z,3+z],1), z=10.);
> > limit(hypergeom([1,2,3,5/2+z], [5/2,2+z,3+z],1), z=100.);
> > limit(hypergeom([1,2,3,5/2+z], [5/2,2+z,3+z],1), z=1000.);
> > limit(hypergeom([1,2,3,5/2+z], [5/2,2+z,3+z],1), z=10000.);
> > limit(hypergeom([1,2,3,5/2+z], [5/2,2+z,3+z],1), z=100000.);
> > limit(hypergeom([1,2,3,5/2+z], [5/2,2+z,3+z],1), z=1000000.);
>
> 1.265299633
> 1.024225676
> 1.002402231
> 1.000240022
> 1.000024000
> 1.000002400
>

Now that I posted, I see that I can do it.
[Written in Maple notation.] Let

> F := (n,x) -> 3/8*GAMMA(n+2)*GAMMA(n+3)*GAMMA(n+5/2+x)*sqrt(Pi)
> *GAMMA(2+x)*GAMMA(3+x)/(GAMMA(5/2+x)*GAMMA(n+5/2)*GAMMA(n+2+x)
> *GAMMA(n+3+x));
F := proc (n, x) options operator, arrow;
3/8*GAMMA(n+2)*GAMMA(n+3)*GAMMA(n+5/2+x)*Pi^(1/2)*GAMMA(2+x)
*GAMMA(3+x)/(GAMMA(5/2+x)*GAMMA(n+5/2)
*GAMMA(n+2+x)*GAMMA(n+3+x)) end proc

We want to compute

> limit(sum(F(n,x),n=0..infinity),x=infinity);
limit(hypergeom([1, 2, 3, 5/2+x], [5/2, 2+x, 3+x], 1), x = infinity)

We claim the limit is 1.
Note that

> simplify(F(x+1,n)-F(x,n));
-3/16*GAMMA(n+3)^2*Pi^(1/2)*n*(13+10*x+2*x^2+5*n+2*x*n)*GAMMA(2+x)^2
*(2+x)*(n+3+x)*GAMMA(n+5/2+x)/((n+2)*GAMMA(n+4+x)^2*GAMMA(7/2+x)
*GAMMA(n+5/2))

is obviously negative for n >=0, x >= 0, so F decreases in x
(at least for integers x).

Next,

> simplify(F(n,2));
72/35*(5+2*n)*(2*n+7)/((n+4)*(n+3)^2*(n+2))

and

> limit(%*n^2,n=infinity);
288/35

so the series sum(F(n,2),n=0..infinity) converges by comparison with
series(1/n^2,n=1..infinity) .

Terms with n >= 1 go to zero:

> limit(F(n,x),x=infinity) assuming n>0;
0

and the term with n=0 is 1:

> simplify(F(0,x));
1

So proceed as follows. Given epsilon > 0, choose N so that
sum(F(n,2),n=N..infinity) < epsilon; and therefore by the decreasing
property, sum(F(n,x),n=N..infinity) < epsilon for all integers x > 2.

Then choose X so that for all x > X,
abs(F(n,x)) < epsilon/N for n=1,2,...,N. Then we get
abs(sum(F(n,x),n=0..infinity) - 1) < 2*epsilon
for all x > X, so the limit is 1 as claimed. 

-- 
G. A. Edgar                               http://www.math.ohio-state.edu/~edgar/