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 (* With[{n = 1000000000}, N[Sum[r^j/(-1 + r^j)^2, {j, 1, Infinity}]/n /. r -> 1 - \[Pi]π/Sqrt[6 n] + \[Pi]^2π^2/(12 n), 10]] *)
 (* 0.9999876725 *)
 Simplify[FindSequenceFunction[{1/2, 5/8, 49/72, 205/288, 5269/7200, 5369/7200, 266681/352800, 1077749/1411200}, k]]
 (* 1/12 (\[Pi]^2π^2 - 6 PolyGamma[1, 1 + k]) *)

 Limit[%, k -> Infinity]
 (* \[Pi]^2π^2/12 *)
 (* With[{n = 1000000000}, N[Sum[r^j/(-1 + r^j)^2, {j, 1, Infinity}]/n /. r -> 1 - \[Pi]/Sqrt[6 n] + \[Pi]^2/(12 n), 10]] *)
 (* 0.9999876725 *)
 Simplify[FindSequenceFunction[{1/2, 5/8, 49/72, 205/288, 5269/7200, 5369/7200, 266681/352800, 1077749/1411200}, k]]
 (* 1/12 (\[Pi]^2 - 6 PolyGamma[1, 1 + k]) *)

 Limit[%, k -> Infinity]
 (* \[Pi]^2/12 *)
 (* With[{n = 1000000000}, N[Sum[r^j/(-1 + r^j)^2, {j, 1, Infinity}]/n /. r -> 1 - π/Sqrt[6 n] + π^2/(12 n), 10]] *)
 (* 0.9999876725 *)
 Simplify[FindSequenceFunction[{1/2, 5/8, 49/72, 205/288, 5269/7200, 5369/7200, 266681/352800, 1077749/1411200}, k]]
 (* 1/12 (π^2 - 6 PolyGamma[1, 1 + k]) *)

 Limit[%, k -> Infinity]
 (* π^2/12 *)
2 added 460 characters in body
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UPDATE

One method how to find this limit is guess a formula for the coefficients. We have

$$r_k=1-\frac{c_k}{\sqrt{n}}+\frac{c_k^2}{2 n}$$

 Simplify[FindSequenceFunction[{1/2, 5/8, 49/72, 205/288, 5269/7200, 5369/7200, 266681/352800, 1077749/1411200}, k]]
 (* 1/12 (\[Pi]^2 - 6 PolyGamma[1, 1 + k]) *)

 Limit[%, k -> Infinity]
 (* \[Pi]^2/12 *)

$$\frac{c_k^2}{2}=\frac{\pi ^2}{12}$$

But such proof is not rigorous...

UPDATE

One method how to find this limit is guess a formula for the coefficients. We have

$$r_k=1-\frac{c_k}{\sqrt{n}}+\frac{c_k^2}{2 n}$$

 Simplify[FindSequenceFunction[{1/2, 5/8, 49/72, 205/288, 5269/7200, 5369/7200, 266681/352800, 1077749/1411200}, k]]
 (* 1/12 (\[Pi]^2 - 6 PolyGamma[1, 1 + k]) *)

 Limit[%, k -> Infinity]
 (* \[Pi]^2/12 *)

$$\frac{c_k^2}{2}=\frac{\pi ^2}{12}$$

But such proof is not rigorous...

1
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Asymptotic solution of the saddle-point equation

As an example we have the following equation:

$$\sum _{j=1}^{\infty } \frac{r^j}{\left(1-r^j\right)^2}=n$$

 Sum[r^j/(1 - r^j)^2, {j, 1, Infinity}] == n

I'm looking for a solution for unknown r. My question is if it possible to find an asymptotic solution of this equation with Mathematica ?

The first iteration (for j = 1) is

 Solve[r/(1-r)^2 == n, r]

 (* {{r -> -((-1 - 2 n + Sqrt[1 + 4 n])/(2 n))}, {r -> (1 + 2 n + Sqrt[1 + 4 n])/(2 n)}} *)

The first of two solutions gives a minimum r

 (* N[% /. n -> 1000] *)
 (* {{r -> 0.968873}, {r -> 1.03213}} *)

Algebraic expressions for next roots are complicated, but here is a numerical approximation:

 (* N[Solve[r/(1 - r)^2 + r^2/(1 - r^2)^2 == n, r] /. n -> 1000] *)
 (* {{r -> -1.01593}, {r -> -0.984316}, {r -> 0.965265}, {r -> 1.03598}} *)

The root with the smallest absolute value is always at position 3. The following program finds an asymptotics iterations from 2 - 5 terms.

 (* Do[so = Solve[(Sum[r^j/(1 - r^j)^2, {j, 1, terms}]) == n, r, Reals]; Quiet[rasy = Table[Expand[FullSimplify[Normal[Series[r /. so[[root]], {n, Infinity, 1}]], n > 0]], {root, 1, Length[so]}]]; Print[rasy[[3]]];, {terms, 2, 5}] *)

 (* 1 - Sqrt[5]/(2 Sqrt[n]) + 5/(8 n) *)
 (* 1 - 7/(6 Sqrt[n]) + 49/(72 n) *)
 (* 1 - Sqrt[205]/(12 Sqrt[n]) + 205/(288 n) *)
 (* 1 - Sqrt[5269]/(60 Sqrt[n]) + 5269/(7200 n) *)

$$1-\frac{\sqrt{5}}{2 \sqrt{n}}+\frac{5}{8 n}$$

$$1-\frac{7}{6 \sqrt{n}}+\frac{49}{72 n}$$

$$1-\frac{\sqrt{205}}{12 \sqrt{n}}+\frac{205}{288 n}$$

$$1-\frac{\sqrt{5269}}{60 \sqrt{n}}+\frac{5269}{7200 n}$$

The final result (the limit of these iterations) is:

$$r \sim 1-\frac{\pi }{\sqrt{6 n}}+\frac{\pi ^2}{12 n}$$

But how to find it with Mathematica ? Generalized methods are welcome.

The numerical check:

 (* With[{n = 1000000000}, N[Sum[r^j/(-1 + r^j)^2, {j, 1, Infinity}]/n /. r -> 1 - \[Pi]/Sqrt[6 n] + \[Pi]^2/(12 n), 10]] *)
 (* 0.9999876725 *)