Double sum with j=i+1 eventual compiling problem

Questions
,
1

Hi all,

can’t make this compile:

Sum[i^(-a)*j^(-a-1),{i,1,n},{j,i+1,n}]

where the real number “a” is supposed to be like: 0 < a < 1.

Wolfram can give answers with unknown numbers, without giving bounds to it.
And this works:

Sum[i^(-2)*j^(-3),{i,1,n},{j,i+1,n}]

There may not be any closed-form of this expression, but in this case, does wolfram give its usual text “Try the following…”?

What could I do?

Thanks in advance.

2

Try with a noninteger value of a

With[{a = 1/2},
 Sum[i^(-a)*j^(-a - 1), {i, 1, n}, {j, i + 1, n}]]

If that gives no closed form, then there is little hope for symbolic a.

3

If I split the sum up, I get only part way:

Sum[j^(-a - 1), {j, i + 1, n}, Assumptions -> 0 < a < 1]
Sum[%, {i, n}, Assumptions -> 0 < a < 1]
(*
HurwitzZeta[1 + a, 1 + i] - HurwitzZeta[1 + a, 1 + n]

Sum[HurwitzZeta[1 + a, 1 + i] - HurwitzZeta[1 + a, 1 + n], {i, n}, 
 Assumptions -> 0 < a < 1]
*)

And one half-step further:

Distribute[%]
(*
Sum[HurwitzZeta[1 + a, 1 + i], {i, n}, Assumptions -> 0 < a < 1] - 
 n Zeta[1 + a, 1 + n]
*)

And that’s as far as I can take it.

4

Thanks guys. I wanted to know if it converges and it does, but I proved it without wolfram…

5

It’s a finite sum, so I’m not sure what you mean by convergence. Have you seen this?:

SumConvergence[i^(-a)*j^(-a - 1), {i, j}]
(*  Re[a] > 1  *)

Perhaps that’s what you wanted.