What is the Fourier series expansion of $\frac{1}{x}$ ?
The best method I could come up with was shifting the function by 'k' (shifting the function to $\frac{1}{x - k}$), so that while calculating the coefficients you don't run into the discontinuity of 1/x.
Is there a different method to calculate the Fourier series of $\frac{1}{x}$.
On
Let $\hat{f}$ be the Fourier transform of some function $f$.
Consider the image of the integer numbers under this Fourier transform $$\{\hat{f}(n) | \ n\in\mathbb{Z}\}$$ Observe that the inverse Fourier series of this set, $$\sum\limits_{n\in\mathbb{Z}} \hat{f}(n) e^{i n}, $$ equals (a.e) some periodic function. If you calculate the Fourier Transform of $1/x$, as it is done here and here, you get the Fourier series coefficients $$c_n = \hat{f}(n).$$ In this case, if the domain of $\frac{1}{x}$ is $[-\pi, \pi)$ then $$c_n = −\frac{i}{2}sgn(n).$$
The same way it must be possible to make the same analysis for the Fourier Series.
On
It is true that $f(x)=1/x$ is not locally integrable at $0$, so there is a problem.
As @reuns points out, once choice is to interpret "$1/x$" as being the Cauchy principal value integral $\lim_{\epsilon\to 0}\int_{\epsilon\le|x|\le\pi}{e^{2\pi inx}\,dx\over x}$. A similar "principal value integral" can be arranged on $[0,2\pi]$, instead, if desired.
But the non-local-integrability cannot be evaded. The "principal value integral" is not a literal integral, since the literal (improper) integral would require that the limits below $0$ and above $0$ be independent... which is not possible.
So, truly, one is probably asking about the Fourier series of a distribution, given by the principal-value-integral (as in @reuns' answer).
The Fourier series only exists for periodic functions which are integrable over a period. You can choose an interval and consider the periodic extension of $\frac{1}{x}$ over that interval, but if that interval contains $0$ (even as an endpoint), it will not be integrable.
At risk of cluttering what was a simple and to-the-point answer, I'd like to address some of the comments this answer has gotten over the years.
It's important to note that, when I say Fourier series, I mean the full Fourier series (not sine or cosine series), and without making use of the Cauchy principal value. For such a series, all the coefficients existing is a necessary condition for the series existing, and since $a_0 \propto \int_L f$, integrability of the function over the chosen period is a necessary condition for the series to exist.
This can be circumvented by using the above modifications. The Cauchy principal value allows odd functions with a singularity at $0$ to retain the property that integrals over a symmetric intervals vanish, so all the $a_n$ coefficients vanish as a result. The sine series, on the other hand, just assumes from the outset that the $a_n$ vanish, with the trade-off of being a series for the odd-periodic extension of $f$.
(Despite the fact that a sine series is a Fourier series, I maintain the distinction in this case because it does not arise from the standard coefficient-based definition of Fourier series without invoking the CPV. You are welcome to disagree, as this is purely a semantic distinction.)
As for square-integrability, that is a sufficient condition, but not necessary for the convergence of a Fourier series, unless you limit "convergence" to mean "convergence in the $L^2$ norm". Different types of convergence have different conditions, and square-integrability is not necessary for all of them; for example, one can construct a non-square-integrable function which has a Fourier series that is pointwise convergent almost everywhere.