1

This is my attempt to prove that Sum 2 also converges. Let Sum 1 i.e. $\sum _{p=0}^{k-1} \left(\frac{\sqrt{\frac{(p+1) \Gamma \left(p+\frac{11}{4}\right)}{\Gamma (p+2)}}}{(p+2) \sqrt{\Gamma \left(\frac{11}{4}\right)}}-\sum _{j=p+2}^k \frac{\sqrt{\frac{\Gamma \left(j+\frac{7}{4}\right)}{\Gamma (j+1)}}}{\sqrt{j} (j+1) \sqrt{\Gamma \left(\frac{11}{4}\right)}}\...


1

Despite your claim, Sum 1 diverges. Here are my arguments. First, Series[Sqrt[Gamma[7/4 + j]/(Gamma[11/4]*Gamma[1 + j])]/(Sqrt[j]*(j+ 1)), {j,Infinity,2}] $$\frac{\left(\frac{1}{j}\right)^{9/8}}{\sqrt{\Gamma \left(\frac{11}{4}\right)}}+O\left(\left(\frac{1}{j}\right)^{17/8}\right) $$ Second, AsymptoticSum[(1/j)^(9/8)/Sqrt[Gamma[11/4]],{j,p+2,k},k -> ...


1

You can try permuting the variables and then sorting and taking the first element of the result. For instance: canon[expr_, vars_List] := First @ Sort @ ReplaceAll[ expr, Thread[vars -> Permutations @ vars] ] Then: c1 = canon[n1 + 2 n2, {n1, n2}] c2 = canon[2 n1 + n2, {n1, n2}] c1 === c2 2 n1 + n2 2 n1 + n2 True


1

Your vars is a list of 40 element, so, given your expression involves quadratic powers of some variables, CoefficientList appears to generate 40-folded list with too many elements (5566277615616, to be precise). So you run out of memory. Meanwhile, most of those elements are zeros (except 36 elements). Use CoefficientRules instead. In particular, Last/@...


Only top voted, non community-wiki answers of a minimum length are eligible