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...