Ignacio Larrosa Caņestro of Coruņa (Spain), Vince Lynch of Doncaster (UK), Jan van Delden of Leeuwarden (The Netherlands), Lou Poulo of Andover MA, Philippe Fondanaiche of Paris (France), Paul Botham of Ipswich (UK), Matt Hudelson of the Washington State University, Robin Stokes of the University of New England (Australia), Nancy Schwarzkopf of New York NY, and Boris Bukh a student at City College of San Francisco solved the problem. Here is Ignacio Larrosa Caņestro's solution using complex analysis followed by Philipe Fondanaiche's solution using differential equations.

**Caņestro's solution:**

For the first question, the coordinates are x = 1 - 1/3 + 1/5 - ... + (-1)^(n-1)/(2n-1) ... = pi/4 from the power series for atan(t), substituting t=1. y = 1/2 - 1/4 + 1/6 - ... + (-1)^(n-1)/(2n) ... = = (1/2)(1 - 1/2 + 1/3 - ... + (-1)^(n-1)/n ...) = (1/2)Ln(2) from the power series for Ln(1 + t), substituting t=1. For the more general question, I will denote angle x by t to avoid confusion with the x-coordinate. Then let z=cos(t) + i*sin(t), with i=sqrt(-1), with t in (-pi, pi]. Then the point P of convergence is P = 1 + z/2 +z^2/3 + ... + z^(n-1)/n + ... ===> z*P = z + z^2/2 + z^3 + ... + z^n/n = - Ln(1 - z), valid for |z|<=1 and z=/=1 (t =/= 0). Then, P = -Ln(1 - z)/z = -Ln(1 - cos(t) - i*sin(t))/(cos(t) + i*sin(t)) = P = -(cos(t) - i*sin(t))Ln(sqrt(2 - 2cos(t))*e^(i*atan(-sin(t)/(1-cos(t))))) P = -(cos(t) - i*sin(t))(Ln(sqrt(2-2cos(t))) - i*((pi/2 - t/2))) P = (pi - t)sin(t)/2 - cos(t)Ln(2 - 2cos(t))/2 + i((pi - t)cos(t)/2 + sin(t)Ln(2 - 2cos(t))/2) ===> P_x = (pi - t)sin(t)/2 - cos(t)Ln(2 - 2cos(t))/2 P_y = (pi - t)cos(t)/2 + sin(t)Ln(2 - 2cos(t))/2 For t = pi/2, P = (pi/4, Ln(2)/2), as we saw previously For t = pi/4, P_x = (3sqrt(2)pi + 4sqrt(2)Ln(1 + sqrt(2)/2))/16 ~= 1.022120903 P_y = (3sqrt(2)pi - 4sqrt(2)Ln(1 + sqrt(2)/2))/16 ~= 0.6439601987 For t=0, the spiral is straight and we get the harmonic series; so there isn't convergence."

**Fondanaiche's solution:**

The coordinates of the point the spiral converges to are defined by the two respective equations: A(x) = 1 + cos(x)/2 + cos(2x)/3 + cos(3x)/4 +...= sum[cos(kx)/(k+1)] for k=0 to infinity B(x) = sin(x)/2 + sin(2x)/3 + sin(3x)/4 +... = sum[sin(kx)/(k+1)] for k=0 to infinity 1) For x=pi/2, we have A(pi/2) = 1-1/3 +1/5 -1/5 +1/9 ..... = pi/4 = 0,78539816... according to the Leibniz formula already analysed in a former SMSU's POTM and B(pi/2) = 1/2 - 1/4 +1/6 ...=Log(2) / 2 = 0,3457359... 2) For x=pi/4 it is relatively easy to find A(pi/4) = sqrt(2)/4*[3*pi/4 + Log(1+sqrt(2)/2)] = 1,0221209.. and B(pi/4) = sqrt(2)/4*[3*pi/4 - Log(1+sqrt(2)/2)] = 0,643960199.. 3) For x=pi/3, we get A(pi/3) = pi*sqrt(3)/6 = 0,906899682.. and B(pi/3) = pi/6 = 0,5236... 4) For x=pi/6, the coordinates of the limit point are A(pi/6) = 5*pi/24 + sqrt(3)/4*Log(sqrt(3)+2) = 1,22475.. and B(pi/6) = 5*pi*sqrt(3)/24 - Log(sqrt(3)+2)/4 = 0,804385.. 5) More generally,by derivating A(x) and B(x) we can write: A'(x) = -sum[k*sin(kx)/(k+1)] and B'(x) = sum[k*cos(kx)/(k+1)] Therefore Y(x) = A(x) + B'(x) = sum[cos(kx)] for k=0 to infinite and Z(x) = B(x) - A'(x) = sum[sin(kx)] for k=0 to infinite Then A(x) - i*A'(x) + B'(x) +i*B(x) = sum[e^ikx] = 1/[1-e^ix] = 1/[1-cos(x) -i*sin(x)] with i^2= -1 and we can infer: Y(x)*(1-cos(x) + Z(x)*sin(x) = 1 Z(x)*(1-cos(x) - Y(x)*sin(x)=0 ==>Y(x)=1/2 and Z(x) = sin(x)/[2*(1-cos(x)] Then we get easily the differential equation of second order: B(x) + B"(x) = sin(x)/[2*(1-cos(x)] defined for x>0 Considering that cos(x) and sin(x) are solutions of B+B"=0 and using the method of variation of parameters with B(x) = u(x)*cos(x) + v(x)*sin(x), we get the solutions u(x)= -(sin(x)+x+C_1)/2 and v(x)=cos(x)/2 +Log(sin(x/2)) + C_2 where C_1 and C_2 are constants Finally the equation of B(x) is: B(x) = (pi-x)*cos(x)/2 +sin(x)*Log(2*sin(x/2)) and the equation of A(x) is A(x) = (pi-x)*sin(x)/2 -cos(x)*Log(2*sin(x/2)) defined for x>0. We can check that both formula are working for x=pi/2, pi/4, pi/3, pi/6 etc...

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