prove the uniform convergence of $\sum_{n=1}^{\infty}\frac{a_n\sin(nx)}{n}$ and $\sum_{n=1}^{\infty}\frac{b_n\cos(nx)}{n}$ Let $f∈L_2[-\pi,\pi]$ and $a_n,b_n (n=1,2,....)$ be a Fourie coefficients of $f$ in the trigonometric system. Prove the uniform convergence of $\sum_{n=1}^{\infty}\frac{a_n\sin(nx)}{n}$ and $\sum_{n=1}^{\infty}\frac{b_n\cos(nx)}{n}$.
Can I just use Bessel's inequality and Weierstrass test? If not, then what should I do? 
 A: The series $\sum a_n^2$ and $\sum b_n^2$ converge by Bessel's inequality for $f \in L_2$, where 
$$\frac{1}{2} a_0^2 + \sum_{n=1}^\infty (a_n^2 + b_n^2) \leqslant \frac{1}{\pi}\|f\|_{L_2}^2$$
By the Cauchy-Schwartz inequality we have
$$\sum_{n=1}^m \frac{|a_n|}{n} \leqslant \left(\sum_{n=1}^m a_n^2 \right)^{1/2}\left(\sum_{n=1}^m \frac{1}{n^2} \right)^{1/2},$$
and the series on the LHS converges since both series on the RHS converge.
Consequently, we can apply the Weierstrass test using
$$\left|\frac{a_n \sin nx}{n}\right| = \frac{|a_n| |\sin nx| }{n} \leqslant \frac{|a_n|  }{n},$$
to prove uniform convergence of $\sum \frac{a_n \sin nx}{n}$, with a similar argument for the other series.
More generally this is true for $f \in L_1$.
While the Fourier series of $f \in L_1$ may or may not converge uniformly, the indefinite integral of $f$ has a uniformly convergent Fourier series.
The function $F(x) = \int_{-\pi}^x f(t) \, dt - (\pi +x)a_0/2$ is continuous and of bounded variation with a convergent Fourier series.
$$F(x) = \frac{1}{2}A_0 + \sum_{n=1}^\infty(A_n \cos nx + B_n \sin nx).$$
The coefficients $A_n, B_n$ can be expressed in terms of the Fourier coefficients $a_n, b_n$ of $f$.
Using integration by parts we find,
$$A_n = \frac{1}{\pi} \int_{-\pi}^\pi F(x) \cos nx \, dx = \left.F(x) \frac{\sin nx}{n}\right|_{-\pi}^{\,\,\pi} - \frac{1}{n\pi}\int_{-\pi}^\pi f(x)\sin nx \, dx = - \frac{b_n}{n}. $$
Similarly, $B_n = a_n/n$, and 
$$F(x) = \frac{1}{2}A_0 + \sum_{n=1}^\infty(\frac{a_n}{n} \sin nx - \frac{b_n}{n} \cos nx ).$$
It is also true that $F$ is absolutely continuous and by a general result the Fourier series, and hence the $\sin$ and $\cos$ component series, are uniformly convergent.
A: The result is not true in general. The Fourier series of a function in $L^2$ converges in $L^2$, but it may fail to converge uniformly. If the convergence were uniforn, the sum would be continuous, and there are plenty of discontinuous functions in $L^2$.
