Something that might prove applicable is the Dirichlet Convergence Test. Summation by Parts yields
$$
\sum_{k=1}^na_kb_k=A_nb_n+\sum_{k=1}^{n-1}A_k(b_k-b_{k+1})\tag{1}
$$
where $\displaystyle A_n=\sum_{k=1}^na_k$. Note that $(-1)^{n+1}\sin(nx)=-\sin(n(\pi+x))$, so we have
$$
\begin{align}
A_n
&=\sum_{k=1}^n(-1)^{n+1}\sin(kx)\\
&=-\sum_{k=1}^n\sin(k(\pi+x))\\
&=\frac{\sin(n(\pi+x)/2)}{\sin((\pi+x)/2)}\sin((n+1)(\pi+x)/2)\tag{2}
\end{align}
$$
Thus, $|A_n|\le|\sec(x/2)|$. Therefore,
(i) according to $(1)$ and $(2)$, the series converges pointwise except at odd multiples of $\pi$; however, at any multiple of $\pi$, each term is $0$. Thus, the series converges for all $x$.
(ii) according to $(1)$ and $(2)$, the series converges uniformly on compact sets not containing an odd multiple of $\pi$. A sequence of continuous functions cannot converge uniformly to a discontinuous function, so the convergence cannot be uniform in any neighborhood of an odd multiple of $\pi$.
(iii) except at multiples of $\pi$, the series is not absolutely convergent. Here is the outline of a proof.
First, show that if $x$ is not a multiple of $\pi$, at least $1/4$ of the multiples of $x$ have $|\sin(kx)|>1/\sqrt{2}$. Because $|\sin(k(x+n\pi))|=|\sin(kx)|$, we only need to consider $x\in[-\pi/2,\pi/2]$. Since $|\sin(-kx)|=|\sin(kx)|$, we only need to consider $x\in[0,\pi/2]$.
If $x\in[\pi/(n+1),\pi/n]$, then at least $\lfloor n/2\rfloor$ multiples of $x$ are in $[(m+1/4)\pi,(m+3/4)\pi]$ out of at most $n+1$ in $[m\pi,(m+1)\pi]$. Since $(0,\pi/2]=\cup_{n=2}^\infty[\pi/(n+1),\pi/n]$, this means that at least $1/4$ of the multiples of $x$ in $[m\pi,(m+1)\pi]$ are in $[(m+1/4)\pi,(m+3/4)\pi]$, and therefore, have $|\sin(kx)|>1/\sqrt{2}$.
Thus, $1/4$ of each $n$ to $n+1$ consecutive terms of the sum have $|\sin(kx)|>1/\sqrt{2}$. The sum is therefore greater than
$$
\frac{1}{4}\sum_{k=n+1}^\infty\frac{2}{k}\frac{1}{\sqrt{2}}\tag{3}
$$
which diverges. Thus, if $x$ is not a multiple of $\pi$, the series of absolute values diverges.
To summarize the apt remarks by Herb Steinberg and Nate Eldridge:
First, let me make clear that I am very sympathetic to this sort of question, as I was very confused by such things when I was learning about Fourier series and related matters. As in "how does the function know that we made it periodic?".
Then the computation of Fourier coefficients depends on the interval we restrict the function to. The corresponding expression in sines and cosines by_its_nature produces a periodic function, which agrees with the original function on the specified interval (more-or-less).
The convergence of the Fourier series to the original function (on the specified interval) is a non-trivial thing in itself. For many reasons, we often want uniform convergence, which is by no means promised. Even raw pointwise convergence is not guaranteed in much generality. Fortunately, these difficulties in our naive conception of "what we want" are not fatal, because it turns out that other forms of convergence (in $L^2$, or in Sobolev spaces) are often sufficient to justify computational conclusions.
Best Answer
See Theorem 2.2 in this pdf which shows (the proof is not really complete) that there is a continuous periodic function with divergent fourier series in some point. The first example of such a function was given by DuBois-Reymond in 1873, in the meantime this has even been extended (see the note on page 13):
Shown in Y. Katznelson, An introduction to harmonic analysis. Wiley 1968.