Is it possible to for the Cesàro means to converge but the Fourier series to diverge?

I was just wondering if it is possible for the Cesàro means to converge but the Fourier series to diverge.

In class we learned that if the Fourier series were to converge, then this limit must equal the limit of the Cesàro means. So, I was just wondering if the converse is also true.

Thanks.

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in general, if $\sum_n a_n$ converges, then $\frac{1}{N}\lim_{n\to\infty}(\sum_{n=1}^Na_n)$ converges. however the converse doesn't hold, eg $a_n=(-1)^n$. you can probably come up with specific examples in fourier series, take the fourier series of bad functions probably – yoyo Apr 8 '11 at 22:18
In general Cesaro convergence does not imply convergence. However it is a theorem of Tauber that if a sequence were $o(1/n)$, then the Cesaro convergence of the corresponding series implies the convergence of the series. This can be applied to functions whose Fourier coefficients $\hat{f}_n$ are such that $n\hat{f}_n \to 0$. – Willie Wong Oct 5 '11 at 15:58

The Cesàro means of a countinuous function $f$ converge uniformly to $f$. But there are continuous functions whose Fourier series diverges at some points (even at a dense set of points). See for instance the answers to this other question. However, if the Fourier series of a continuous function $f$ converges at a point $x$, then its sum is $f(x)$.