This question was originally about evaluating the sum $\sum_{n=0}^\infty e^{nix}$, but I figured out the answer about half way through writing it. So instead, I decided to ask a slightly different question.
Now, obviously, the sum $\sum_{n=0}^\infty e^{nix}$ does not converge – however, the closure of the set of points $(x,z)$ given by $z=\sum_{n=0}^\omega e^{nix}$ is bounded for arbitrarily large $\omega$; the object formed by these points is well defined as $\omega\to\infty$.
This becomes clear when you break the sum into a real and imaginary part:
$$\sum_{n=0}^\infty e^{nix}=\sum_{n=0}^\infty \cos{nx}+i\sum_{n=0}^\infty \sin{nx}$$
Both sums form bounded sets, and the boundary is clearly another periodic function with vertical asymptotes at integer multiples of $2\pi$.
The resulting object can be expressed (admittedly awkwardly) as a piece-wise set-valued function of a single real variable $x$:
$$r_1(x)=-\frac{1}{2}\csc{\frac{x}{2}}+\frac{1}{2}\qquad r_2(x)=\frac{1}{2}\csc{\frac{x}{2}}+\frac{1}{2}$$
$$m_1(x)=-\frac{1}{2}\tan{\frac{x}{4}}\qquad m_2(x)=\frac{1}{2}\cot{\frac{x}{4}}$$
$$f(x)=\begin{cases}[r_1(x),r_2(x)]+i[m_1(x),m_2(x)]&\cot{\frac{x}{4}}>0\\ \mathbb{C}&\cot{\frac{x}{4}}=0^{\pm1}\\ [r_2(x),r_1(x)]+i[m_2(x),m_1(x)]&\cot{\frac{x}{4}}<0\end{cases}$$
Using $0^{-1}=\pm\infty$, $\quad Y_1+Y_2=\left\{y_1+y_2\mid y_1\in Y_1\land y_2\in Y_2\right\}$, and $iY=\left\{iy\mid y\in Y\right\}$. (If there is a more elegant way to write this, please tell me).
Naturally my next question was whether or not other 'spiky summations' exist, so I played around with different periodic functions, trying to get the summation to 'converge' to a bounded shape. After experimenting for a while, it seems that there is an entire class of these objects – which have interesting geometric and statistical properties.
Do these objects have a name? And is there a general method for finding the bounding curves given the summation used to generate them?
Best Answer
This behavior is very easy to explain when you view the sum as a geometric series. We have a closed formula $$\sum_{n=0}^\omega e^{nix}=\frac{e^{(\omega+1)ix}-1}{e^{ix}-1}$$ and the bounding curves you are finding are just what you get by maximizing and minimizing the real and imaginary parts of $$\frac{w-1}{e^{ix}-1}$$ where $w$ is allowed to range over the unit circle. When $\omega$ is large, $e^{(\omega+1)ix}$ is cycling through the entire unit circle very quickly as $x$ changes (while the denominator $e^{ix}-1$ is changing slowly), and so the real and imaginary parts of the sum quickly oscillate between the maximum and minimum values where $w$ ranges over the unit circle.
Explicitly, to find the real and imaginary parts we write $w=a+bi$ and find $$\frac{w-1}{e^{ix}-1}=\frac{a-1+bi}{\cos x-1+i\sin x}=\frac{a(\cos x-1)+b\sin x+1-\cos x}{(\cos x-1)^2+\sin^2 x}+i\frac{-a\sin x+b(\cos x-1)+\sin x}{(\cos x-1)^2+\sin^2 x}.$$ By Cauchy-Schwarz, given the constraint $a^2+b^2=1$, we maximize or minimize $a(\cos x-1)+b\sin x$ by making $(a,b)$ a scalar multiple of $(\cos x-1,\sin x)$, so that $a=\pm\frac{\cos x-1}{\sqrt{(\cos x-1)^2+\sin^2 x}}$ and $b=\pm\frac{\sin x}{\sqrt{(\cos x-1)^2+\sin^2 x}}$. This tells us that when $2-2\cos x$ is positive, the real part is bounded by $$\frac{1}{2}\pm\frac{(\cos x-1)^2+\sin^2 x}{((\cos x-1)^2+\sin^2 x)^{3/2}}=\frac{1}{2}\pm\frac{1}{2}\csc(x/2).$$ Similarly, the imaginary part is bounded by $$\frac{1}{2}\frac{\sin x}{1-\cos x}\pm\frac{1}{2}\csc(x/2)$$ which is $\frac{1}{2}\cot(x/4)$ or $-\frac{1}{2}\tan(x/4)$ depending on the sign chosen.
The tangent-like appearance of the imaginary part graph (with half the period of the bounding curves) is explained by the fact that the average value between the oscillations is $\frac{1}{2}\frac{\sin x}{1-\cos x}=\frac{1}{2}\cot(x/2)$. In other words, the imaginary part is the graph of $\frac{1}{2}\cot(x/2)$ superimposed with oscillations whose peaks behave like $\frac{1}{2}\csc(x/2)$.
The bounding curves for other similar summations can be found in the same way, by writing everything in terms of complex exponentials and then summing as a geometric series. In particular, the same method should work for three of the four examples you presented at the end (the exception being $\sum (-1)^n \cos^2(n/(x-n))$, which is really not the same sort of sum at all and has very different behavior).