Could you please help me to clarify the following concepts:
1) A Fourier series $\sum_{-\infty}^\infty a_n e^{inx}$ is given by specifying the coefficients $a_n$ in some way. Why the condition $\sum_{-\infty}^\infty |a_n|<\infty$ is necessary to be sure that the series converges to a continuous function?
2) Why $\sum_{-\infty}^\infty |a_n|^2<\infty$ is a weaker condition than $\sum_{-\infty}^\infty |a_n|<\infty$? (because $a_n$ are less than 1?)
On
If $\sum_{n}|a_n| < \infty$, then $\sum_n a_n e^{inx}$ converges uniformly by the Weierstrass M-test, which means that the series $s(x)=\sum_{n}a_n e^{inx}$ is a continuous function. (However, it is not necessary for this condition to hold in order for the series to converge to a continuous function.) Because $s(x)$ is continuous, it is in $L^2[0,2\pi]$, which also gives the following Parseval identity: $$ \frac{1}{2\pi}\int_{0}^{2\pi}|s(x)|^2 dx = \sum_{n}|a_n|^2 < \infty. $$ Therefore $\sum_n |a_n| < \infty \implies \sum_n |a_n|^2 < \infty$, which proves that the second condition is weaker.
Condition 1) just ensures that the series converges.
Condition 2) is more general. For example, the sequence $a_n = {1 \over n}$ is square summable but not summable, and all summable sequences are square summable since if $|a_n| \le 1$ then $|a_n|^2 \le |a_n|$.
Addendum:
The above 1) doesn't actually answer what the OP asked. The absolute summability of $a_n$ is sufficient for the (non symmetric) partial sums to converge to a continuous function.