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.
 A: A standard example is 
$$
f(t)= \sum_{n>1} \frac{\sin(nt)}{\log(n)}$$
The conjugate of $f$ is a Fourier series but $f\not\in L^1(\mathbb{T})$ and hence is no Fourier series.
For further explanation see Katznelson's book page 85.
(Edit: If $f$ is not in $L^1(\mathbb{T})$ it is hard to define Fourier coefficients.
Added 29/8 - 2010
Here is a screen dump from Zygmund's book "Trigonometric Series Vol I"
.
However, in order to be a Fourier series the coefficients are to be Fourier coefficients of some function $f$, these are well defined if $f\in L^1(\mathbb{T})$ (because these are calculated using integration against a bounded measurable function). Thus it is natural to define a trigonometric series if the coefficients are Fourier coefficients of some $L^1$ function. Hence, all answers given to this problem are in fact right! 
Moreover, the Riemann-Lebesgue Lemma states that the Fourier coefficients of an $L^1$-function tends to $0$ as the index approach $\pm\infty$. Thus, it is easy to construct formal trigonometric series that are not Fourier series.
The example $f$ given above does in fact converge everywhere and hence this is a kind of non-trivial example of a trigonometric series that is not a Fourier series.
A: Theorem. If $a_n>0$ and $\sum_{n>0}\frac{a_n}{n}=\infty$. Then $\sum_1^\infty a_n\sin nt$ is not a Fourier series. (AN INTRODUCTION TO HARMONIC ANALYSIS, Yitzhak Katznelson)
A: 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.  
A: When all the coefficients are 1, the series does not converge to a function.  (It does converge in the sense of a distribution, to the Dirac Delta.).
A: Using Riemann-Lebesgue lemma we can give the example of a trigonometric series which is not a Fourier series.
e.g.$\sum_{n=1}^{\infty}\cos nx$ is not a Fourier series as its coefficient does not tends to zero.
For better understand on this topics i would suggest you to prefer Bary( Vol.1) book.
A: 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.
