5
$\begingroup$

I want to check that the following series $$ s_1=\sum_{i,j=1}^{\infty}\frac{ij}{(i+j)^{a/2+1}} $$ diverges for $a\le 6$, where $a$ is a positive integer. I tried with SumConvergence: SumConvergence[(i j)/(i + j)^(a/2 + 1), {i, j}]without getting an answer. Can anyone help?

$\endgroup$

3 Answers 3

5
$\begingroup$

The simplest approach is to replace the sum by an integral:

Integrate[(x y)/(x + y)^(a/2 + 1), {x, 1, ∞}, {y, 1, ∞}]

(*... Re[a] > 6 *)
$\endgroup$
7
$\begingroup$

For greater clarity, I will consider a series

 Sum[(i j)/(i + j)^k, {i, 1, Infinity}]

Mathematica does not generally sum this series, but we can derive a general formula

 Table[Sum[(i j)/(i + j)^k, {i, 1, Infinity}], {k, 4, 10}]
 (* {-(1/6) j (3 PolyGamma[2, 1 + j] + j PolyGamma[3, 1 + j]), 
    1/24 j (4 PolyGamma[3, 1 + j] + j PolyGamma[4, 1 + j]),
    -(1/120) j (5 PolyGamma[4, 1 + j] + j PolyGamma[5, 1 + j]), 
    1/720 j (6 PolyGamma[5, 1 + j] + j PolyGamma[6, 1 + j]),
    -((j (7 PolyGamma[6, 1 + j] + j PolyGamma[7, 1 + j]))/5040),
    (j (8 PolyGamma[7, 1 + j] + j PolyGamma[8, 1 + j]))/40320,
    -((j (9 PolyGamma[8, 1 + j] + j PolyGamma[9, 1 + j]))/362880)} *)

From here we have the same result

 Table[(-1)^(k + 1) 1/(k - 1)! j ((k - 1)*PolyGamma[k - 2, 1 + j]
 + j PolyGamma[k - 1, 1 + j]), {k, 4, 10}]

Now we will see how these expressions behave for large j

 Table[Series[(-1)^(k + 1) 1/(k - 1)! j ((k - 1)*
  PolyGamma[k - 2, 1 + j] + j PolyGamma[k - 1, 1 + j]), {j, 
  Infinity, k}], {k, 4, 10}]
 (* {1/(6 j)-1/(12 j^3)+O((1/j)^5),
    1/(12 j^2)-1/(12 j^4)+O((1/j)^6),
    1/(20 j^3)-1/(12 j^5)+O((1/j)^7),
    1/(30 j^4)-1/(12 j^6)+O((1/j)^8),
    1/(42 j^5)-1/(12 j^7)+O((1/j)^9),
    1/(56 j^6)-1/(12 j^8)+O((1/j)^10),
    1/(72 j^7)-1/(12 j^9)+O((1/j)^11)} *)

The terms with the highest weight are

 Table[1/((k - 1)*(k - 2)*j^(k - 3)), {k, 4, 10}]
 (* {1/(6 j), 1/(12 j^2), 1/(20 j^3), 1/(30 j^4), 1/(42 j^5), 1/(56 j^6), 1/(72 j^7)} *)

The sum_{j=1..infinity} 1/j^n converges for n>1, so we have

  k - 3 > 1 /. k -> a/2 + 1 // Simplify
  (* a > 6 *)
$\endgroup$
4
$\begingroup$
$Version

(* "13.3.1 for Mac OS X ARM (64-bit) (July 24, 2023)" *)

Clear["Global`*"]

sum[α_] := sum[α] =
  Sum[i*j/(i + j)^(α/2 + 1), {i, 1, ∞}, {j, 1, ∞}]

Table[{α, sum[α]}, {α, 5, 10}]

enter image description here

sum is only defined for even values greater than 6

data = Table[{α, sum[α]}, {α, 8, 30, 2}]

(* {{8, -(1/540) π^2 (-15 + π^2)}, {10, 1/6 (Zeta[3] - Zeta[5])}, {12, (
  21 π^4 - 2 π^6)/11340}, {14, 
  1/6 (Zeta[5] - Zeta[7])}, {16, -((π^6 (-10 + π^2))/56700)}, {18, 
  1/6 (Zeta[7] - Zeta[9])}, {20, (π^8 (99 - 10 π^2))/5613300}, {22, 
  1/6 (Zeta[9] - Zeta[11])}, {24, (π^10 (6825 - 691 π^2))/
  3831077250}, {26, 
  1/6 (Zeta[11] - Zeta[13])}, {28, (π^12 (691 - 70 π^2))/
  3831077250}, {30, 1/6 (Zeta[13] - Zeta[15])}} *)

For even values the sums are

f[α_] = 1/6 (Zeta[(α - 4)/2] - Zeta[α/2]);

Verifying the sums for even values,

And @@ Table[sum[α] == f[α], {α, 8, 30, 2}]

(* True *)

f has a singularities at α == 2 and α == 6

FunctionDomain[f[α], α]

(* α < 2 || 2 < α < 6 || α > 6 *)

Plot[f[α], {α, 0, 10}, PlotRange -> All]

enter image description here

Show[
 Plot[f[α], {α, 7, 30},
  PlotRange -> All],
 DiscretePlot[sum[α], {α, 8, 30, 2},
  PlotStyle -> Red]]

enter image description here

Show[
 LogPlot[f[α], {α, 7, 30},
  PlotRange -> All],
 DiscretePlot[sum[α], {α, 8, 30, 2},
  PlotStyle -> Red,
  ScalingFunctions -> "Log"]]

enter image description here

Use f for the sum, it is much faster

sum[48] // AbsoluteTiming

(* {26.9225, (π^22 (2332820490 - 236364091 π^2))/1211517431782539131250} *)

f[48] // AbsoluteTiming

(* {0.000031, 
 1/6 ((155366 π^22)/13447856940643125 - (236364091 π^24)/
    201919571963756521875)} *)

%[[2]] == %%[[2]] // Simplify

(* True *)
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.