For what $n$ is $\sum_{i=1}^\infty \frac{\cos (it)}{i^n}$ bounded and why doesn't a sine behave the same way? 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!
 A: 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
$$
\sum_{k=1}^\infty\frac{\sin(kt)}{k^n}=\Im\left(\Li_n\left(e^{it}\right)\right)\tag{3}
$$

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
$$
\lim_{t\to0}\left|\sum_{k=1}^\infty\frac{\sin(kt)}k\right|=\frac\pi2\tag{6}
$$

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
$$
\lim_{t\to0}\sum_{k=1}^\infty\frac{\sin(kt)}{k^2}=0\tag{10}
$$

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