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 - π/Sqrt[6 n] + π^2/(12 n), 10]] *)
(* 0.9999876725 *)
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 (π^2 - 6 PolyGamma[1, 1 + k]) *)
Limit[%, k -> Infinity]
(* π^2/12 *)
$$\frac{c_k^2}{2}=\frac{\pi ^2}{12}$$
But such proof is not rigorous...