Three Sums

The given equation can be rewritten as the quartic $r_kx^4 - 4r_kx^3 + (4r_k - 2)x^2 + 4x - 4 = 0,$ where $r_k = k(k + 1)$ , and this can be factored as $\left( x^2 - 2x - \frac 2k \right)(r_kx^2 - 2r_kx + 2k) = 0.$ The solutions obtained from the second quadratic factor turn out to be extraneous. The solutions from the first quadratic factor are $x = 1 \pm \sqrt{1 + \frac 2k} \implies x_k = 1 + \sqrt{1 + \frac 2k}.$ Using this expression for $x_k$ , we can compute all the values of $S_n$ , $T_n$ , and $U_n$ directly (e.g., using a script like the one below). When $n = 239$ , we have $U_n \approx 200100.099$ , and when $n = 240$ , we have $U_n \approx 202172.735$ . Therefore, the desired value of $n$ is $\boxed{240}$ .

def x(n):
    return 1 + (1 + 2/n)**(1/2)
def s(n):
    if n == 1:
        return 1/(2 + 2*x(1))**(1/2)
    return s(n - 1) + 1/(2 + 2*x(n))**(1/2)
def t(n):
    if n == 1:
        return 1/(1 + 2**(1/2) + 2*s(1))
    return t(n - 1) + 1/(1 + 2**(1/2) + 2*s(n))
def u(n):
    if n == 1:
        return t(1)**2 + 1/t(1)**2
    return u(n - 1) + t(n)**2 + n**2/t(n)**2
print(u(239))
print(u(240))