Example of a trigonometric series that is not fourier series? My textbook doesn't give any example of this kind of series. Could you provide some?

Trigonometric series is defined in wikipedia as :
$A_{0}+\sum_{n=1}^{\infty}(A_{n} \cos{nx} + B_{n} \sin{nx})$

When 

$A_{n}=\frac{1}{\pi} \int^{2 \pi}_0 f(x) \cos{nx} dx\qquad (n=0,1,2,3 \dots)$

$B_{n}=\frac{1}{\pi} \int^{2 \pi}_0 f(x) \sin{nx} dx\qquad (n=1,2,3, \dots)$

It is fourier series.

thanks.
I don't know the answer (I should have! -- see below), but in my opinion a lot of people are misinterpreting the question, so perhaps it is worth an answer to try to set this straight.

Here is an analogy: trigonometric series is to Fourier series as power series is to Taylor series.  In other words, a trigonometric series is just any series of the form

$A_{0}+\sum_{n=1}^{\infty}(A_{n} \cos{nx} + B_{n} \sin{nx})$

This should be understood formally, i.e., this is a trigonometric series even if it doesn't converge anywhere.  A fourier series is the trigonometric series associated to an 
$L^1$ function by taking $A_n$ and $B_n$ as above.

Just as a Taylor series need not need converge at any point except the central point, a Fourier series need not converge pointwise at any point.  Thus the Fourier series need not be a function and in particular the Fourier inversion theorem need not apply.  

A theorem of Borel asserts that given any sequence $(a_n)$ of real numbers, there exists a $C^{\infty}$-function on the real line whose Taylor series at $0$ is $\sum_{n=0}^{\infty} a_n x^n$.  In particular, if the $a_n$'s grow too rapidly, the Taylor series will diverge away from zero and the function $f$ will not be analytic.  

I interpret the question as asking whether the analogue of Borel's theorem is true for Fourier series: is every trigonometric series the Fourier series of some $L^1$ function (even if the trigonometric series does not converge to the function)?  

P.S.: Boo to wikipedia for asserting the answer to this question without giving a reference.

Addendum: As Pierre-Yves Gaillard points out, the Fourier coefficients of any $L^1$ function $f$ are uniformly bounded (by $||f||_1$), so this answers the question as I have interpreted it.
Take a look at this wikipedia stub. A trigonometric series, which is not a Fourier series, must therefore be one that is defined by

$$ f(x) = A_{0}+\displaystyle\sum_{n=1}^{\infty}(A_{n} \cos{nx} + B_{n} \sin{nx}) $$

for which the relations

$$ A_{0}=\frac{1}{2\pi}\displaystyle\int^{2 \pi}_0 f(x)\,dx $$

$$ A_{n}=\frac{1}{\pi}\displaystyle\int^{2 \pi}_0 f(x) \cos{nx} \,dx\qquad (n=1,2,3 \dots)$$

$$B_{n}=\frac{1}{\pi}\displaystyle\int^{2 \pi}_0 f(x) \sin{nx}\, dx\qquad (n=0,1,2,3, \dots)$$

do not hold. But since {$\sin(nx),\cos(nx)$} forms a complete orthonormal basis, I cannot imagine one. Except for a non-convergent one perhaps...

edit As J. Mangaldan pointed out: Lacunary trigonometric series might be the answer you search, i.e. series that cannot be analytically be continued.