Sometimes substitution of hgfred for %f will simplify an expression. But not for this case:
(%i29) simplify_sum(sum(x^(3*n)/(3*n)!,n,0,inf));
(%o29) %f[0,2]([],[1/3,2/3],x^3/27)
(%i30) subst(%f[0,2] = hgfred,%);
(%o30) hgfred([],[1/3,2/3],x^3/27)
Oops--right back where we started from:
(%i31) ev(%,nouns);
(%o31) %f[0,2]([],[1/3,2/3],x^3/27)
And for the record, I think the simplify_sum gives a correct value (but maybe not what is desired):
(%i27) simplify_sum(sum(x^(3*n)/(3*n)!,n,0,inf));
(%o27) %f[0,2]([],[1/3,2/3],x^3/27)
(%i28) taylor(subst(%f[0,2] = hypergeometric,%),x,0,19) - sum(x^(3*n)/(3*n)!,n,0,6);
(%o28)/T/ 0+...
--Barton
> Unfortunately, in your case, simplify_sum gives the result as the hypergeometric function %f[0,2]([],[1/3,2/3],x^3/27), which may be correct, but isn't very helpful.