Infinite Arctan Series?

I am sorry, I just discovered a sign error for (a_n)² in the previous version, the coefficients were also wrong. Here is the corrected version.

As already corrected, the statement to prove is

π/12 = Σ {from 1 to ∞} arctan[1/(a_n)²]

Obviously, one can replace arctan[1/(a_n)²] by arccot[(a_n)²].

We first solve a_n. The eigen equation is

λ² = 4 λ −1

The equation essential asks the following question: assume a_n = λⁿ, what is λ?

Its solution is λ = 2 ±√3.

The general solution of a_n can be written as

a_n = C1(2+√3)ⁿ − C2 (2−√3)ⁿ

using the two initial conditions to determine C1 and C2, we have

a_n = [(2+√3)ⁿ − (2−√3)ⁿ]/√3

Thus, (a_n)² can be expressed as

(a_n)² = (1 − βⁿ)² / (3 βⁿ)

where β = (2−√3)²

Now let us define a partial sum as

T_n = tan{ Σ {from 1 to n} arctan[1/(a_k)²] }

obviously T_n satisfies,

arctan(T_n) − arctan(T_{n−1}) = arctan[1/(a_n)²]

Now, take the tangent of both sides,

(a) [T_n −T_{n−1}]/[1+ T_n T_{n−1}] = 1/(a_n)² = (3 βⁿ)/(1 − βⁿ)²

We now wish to find a solution of T_n.

Since the right hand side is a function of βⁿ, somust be T_{n−1} and T_n.

Our guess is T_{n−1} = [u+v βⁿ] / [1+w βⁿ], and T_n = [u+v β βⁿ] / [1+w β βⁿ]. Plug it into the above formula and determine the coefficients u,v,w. Also notice that, the initial condition T_0 = 0 requires that u + v β = 0.

It turns out this guess is correct (there is a set of u,v,w that satisfies equation (a) and the initial condition). By expand (a) and compare coefficients, we have:

T_{n−1} = [2−√3 − (2+√3) βⁿ] / (1− βⁿ)

or

T_n = (2−√3) [ 1− βⁿ] / (1− β βⁿ)

Now take the limit n -> ∞, we have, T_∞ = 2−√3.

arctan(T_∞) = π/12.

clearly a_n is increasing with n.

[in details: if a(n-1) > a(n-2)>0, then a(n-1) - a(n-2) > 0, and

a(n) = 4a(n-1) - a(n-2) = 3a(n-1) + a(n-1) - a(n-2) > 3a(n-1) > a(n-1).

So a(n) is increasing when the first two terms a(1) and a(2) were given as increasing.]

and so arctan[(a_n)^2] does not tend to zero as n->infinity (in fact you can show that a_n -> infinity, so arctan[(a_n)^2] -> pi/2)

and the series doesn't converge.

maybe you made a typo.