MHB Destination Points in Harmonic Sequences

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The discussion centers on the convergence of destination points in harmonic sequences for values of θ within the range 0 ≤ θ < 2π. It is established that for 0 < θ < 2π, a limit point can be reached, which can be calculated using complex numbers and the series sum. The limit point is expressed in terms of coordinates involving logarithmic functions and trigonometric identities. The graph of these limit points is provided, illustrating specific values such as θ = π/2 and θ = π, which correspond to known series results. The complexity of finding a function that represents this set of points is acknowledged, highlighting the intricate nature of the problem.
wheepep
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The pattern above will continue for all values of the harmonic sequence.

Will a destination point be reached for any value of θ where 0 ≤ θ < 2𝜋?

(I know it won’t for θ = 0)

Is there a function which contains the set of all destination points?
 

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wheepep said:
The pattern above will continue for all values of the harmonic sequence.

Will a destination point be reached for any value of θ where 0 ≤ θ < 2𝜋?

(I know it won’t for θ = 0)

Is there a function which contains the set of all destination points?
The answer is Yes. If $0<\theta<2\pi$ then the path will converge to a limit point, which you can find by using complex numbers.

The limit point will be the sum of the series $1 + \tfrac12e^{i\theta} + \tfrac13e^{2i\theta} + \tfrac14e^{3i\theta} + \ldots$. You can sum that series using the power series expansion $\log(1-x) = -x - \frac12x^2 - \frac13x^3 - \ldots$, like this: $$\begin{aligned}1 + \tfrac12e^{i\theta} + \tfrac13e^{2i\theta} + \tfrac14e^{3i\theta} + \ldots &= e^{-i\theta}\sum_1^\infty\frac{e^{in\theta}}n \\ &= -e^{-i\theta}\log\bigl(1 - e^{i\theta}\bigr) \\ &= -e^{-i\theta}\log(1 - \cos\theta - i\sin\theta) \\ &= -e^{-i\theta}\log\bigl( 2\sin^2\tfrac\theta2 - 2i\sin\tfrac\theta2\cos\tfrac\theta2 \bigr) \\ &= -e^{-i\theta}\log\Bigl(2\sin\tfrac\theta2 \cdot e^{i(\theta - \pi)/2}\Bigr) \\ &= (-\cos\theta + i\sin\theta) \bigl(\log \bigl( 2\sin\tfrac\theta2\bigr) + i\tfrac12(\theta - \pi)\bigr) \end{aligned}$$ In terms of coordinates, this shows that your "destination point" is the point $$\Bigl(-\cos\theta\log \bigl( 2\sin\tfrac\theta2\bigr) + \tfrac12(\pi - \theta)\sin\theta\,,\, \sin\theta\log \bigl( 2\sin\tfrac\theta2\bigr) + \tfrac12(\pi - \theta)\cos\theta \Bigr).$$ That looks complicated. But I think it is probably correct because it gives the right result $(\log2,0)$ when $\theta = \pi$. (In that case, the series just becomes $1-\frac12 + \frac13 - \ldots$, which converges to $\log2$ on the real axis.)

If you are asking for a function that has this set of points as its graph, I think you can see that this would be a pretty hard problem. (Worried)
 
That's great. Thanks!
 
Hey wheepep! ;)

Here's the graph of the limit points based on Opalg's formula:

\begin{tikzpicture}
%preamble \usepackage{pgfplots}
\begin{axis}[xmin=0, grid=both, axis lines=middle, axis equal, xlabel=$x$, ylabel=$y$]
\addplot[ultra thick, red, domain=0:3.142, variable=t, samples=51]
( {-cos(deg(t))*ln(2*sin(deg(t/2)))+1/2*(pi-t)*sin(deg(t))},
{sin(deg(t))*ln(2*sin(deg(t/2)))+1/2*(pi-t)*cos(deg(t))} );
\addplot[red, mark=ball] coordinates {({pi/4}, {ln(2)/2})} node[above left] {$\theta=\pi/2$};
\addplot[red, mark=ball] coordinates {(ln(2), 0)} node[above left] {$\theta=\pi$};
\end{axis}
\end{tikzpicture}

In particular we can see that Leibniz series is also in there for $\theta=\frac\pi 2$ with $x=1-\frac 13 + \frac 15 - ... =\frac \pi 4$.
That limit point has $y=\frac 12 - \frac 14 + \frac 16 - ... = \frac 12(1-\frac 12 +\frac 13 - ...)= \frac 12 \ln 2$, which is half of the Alternating harmonic series.
 
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