For what $n$ is $\sum_{i=1}^\infty \frac{\cos (it)}{i^n}$ bounded and why doesn't a sine behave the same way?

Question

I've been looking at a parametric curve $$\pmatrix{X \\ Y}=\pmatrix{\sum_{i=1}^N \frac{\cos (it)}{i^n} \\ \sum_{i=1}^N \frac{\sin (it)}{i^n}}$$ where, for the plots below, $N$ runs from $1 \rightarrow 300$ and $n=1,2$, respectively.

It seems that $X$ is unbounded/divergent in one, and bounded/convergent in the other, whereas $Y$ seems to be indifferent to $n$ and always be bounded.
I use the term "bounded" to encapsulate that several different properties ($ \max (X),$ the area enclosed by $X,...$) could be a measure of this.

My question: For which $n \in \mathbb{R}$ is $X,Y$ bounded, and why does the sine always seem to be bounded?

I guess it could have something to do with this post, but I don't quite see how the argument in that post would be used for this problem, mostly because of my $t$, but perhaps it doesn't change anything?

As a sidenote, I tried introducing a $(-1)^{i+1}$ in the sums, but all this did was to mirror the graph around the left-most point of the original graph, so that $X$ diverged to $-\infty$ instead. Any ideas of why this is the case?

$n=1$

$n=2$

Oh, and here is a nice and wobbly version with $(-1)^{i}$:

$n=1$

Any insights are much appreciated!

Answer 1

The function we should look at is called the Polylogarithm function $$ \newcommand{\Li}{\operatorname{Li}} \newcommand{\Re}{\operatorname{Re}} \newcommand{\Im}{\operatorname{Im}} \newcommand{\sign}{\operatorname{sign}} \sum_{k=1}^\infty\frac{x^k}{k^n}=\Li_n(x)\tag{1} $$ Then, your functions are $$ \sum_{k=1}^\infty\frac{\cos(kt)}{k^n}=\Re\left(\Li_n\left(e^{it}\right)\right)\tag{2} $$ and $$ \sum_{k=1}^\infty\frac{\sin(kt)}{k^n}=\Im\left(\Li_n\left(e^{it}\right)\right)\tag{3} $$

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For $n=1$, we have $$ \begin{align} \sum_{k=1}^\infty\frac{e^{ikt}}k &=-\log\left(1-e^{it}\right)\\ &=-\log(2-2\cos(t))+i\sign(t)\left(\frac\pi2-\frac{|t|}2\right)\tag{4} \end{align} $$ The real part of $(4)$ says $$ \lim_{t\to0}\sum_{k=1}^\infty\frac{\cos(kt)}k=\infty\tag{5} $$ and the imaginary part of $(4)$ says $$ \lim_{t\to0}\left|\sum_{k=1}^\infty\frac{\sin(kt)}k\right|=\frac\pi2\tag{6} $$

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For $n=2$, we have $$ \sum_{k=1}^\infty\frac1{k^2}=\frac{\pi^2}6\tag{7} $$ Therefore, by Dominated Convergence (which is valid for infinite sums using a discrete measure), $$ \begin{align} \lim_{t\to0}\sum_{k=1}^\infty\frac{e^{ikt}}{k^2} &=\frac{\pi^2}6\tag{8} \end{align} $$ The real part of $(8)$ is $$ \lim_{t\to0}\sum_{k=1}^\infty\frac{\cos(kt)}{k^2}=\frac{\pi^2}6\tag{9} $$ and the imaginary part of $(8)$ is $$ \lim_{t\to0}\sum_{k=1}^\infty\frac{\sin(kt)}{k^2}=0\tag{10} $$

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The case for any real $n\gt1$ is similar to $n=2$ since for $n\gt1$, $$ \sum_{k=1}^\infty\frac1{k^n}=\zeta(n)\lt\infty\tag{11} $$

Answer 2

This is too long for a comment, and a slightly different view-point than the excellent answer by robjohn, for the ones not knowing non-elementary functions:

For $n>1$, the series $\sum_{k=1}^{+\infty}\frac{\cos(kt)}{k^n}$ and $\sum_{k=1}^{+\infty}\frac{\sin(kt)}{k^n}$ are uniformly convergent (by Weierstrass M-test since the series $\sum_{k=1}^{+\infty}1/k^n$ converges for $n>1$.

Since we have continuous functions converging uniformly we know that the limits are continuous functions, and we should not be surprised of the boundedness.

The case $0<n\leq 1$ is more interesting. In this case we can use Dirichlet's test for convergence.

Due to periodicity, it is sufficient to consider $0\leq t<2\pi$. From the well-known formulas (these can be proven using geometric sums) $$ \sum_{k=1}^N\cos(kt)=\cos\bigl((N+1)t/2\bigr)\frac{\sin(Nt/2)}{\sin(t/2)} $$ and $$ \sum_{k=1}^N\sin(kt)=\sin\bigl((N +1)t/2\bigr)\frac{\sin(Nt/2)}{\sin(t/2)} $$ we find that, at least for fixed $t$ in $0<t<2\pi$, $$ \Bigl|\sum_{k=1}^N\cos(kt)\Bigr|\leq\frac{1}{\sin(t/2)} $$ and $$ \Bigl|\sum_{k=1}^N\sin(kt)\Bigr|\leq\frac{1}{\sin(t/2)} $$ uniformly in $N$. Moreover, the sequence $k\mapsto 1/k^n$ is clearly decreasing in $k$ and has limit $0$ as $k\to+\infty$. Dirichlet's test applies, and we find that the series $$ \sum_{k=1}^{+\infty}\frac{\cos(kt)}{k^n}\quad\text{and}\quad \sum_{k=1}^{+\infty}\frac{\sin(kt)}{k^n} $$ converge for $0<n\leq 1$ and $0<t<2\pi$ and, thus, for each $t$, the partial sums are bounded.

The case $t=0$ remains. But for $t=0$, we have $$ \sum_{k=1}^N\frac{\cos(kt)}{k^n}=\sum_{k=1}^N\frac{1}{k^n} $$ and $$ \sum_{k=1}^N\frac{\sin(kt)}{k^n}=\sum_{k=1}^N 0=0. $$

The first sum tends to $+\infty$ as $N\to+\infty$ and the second sum clearly converges to $0$ s $N\to+\infty$.