Absolute and Uniform Convergence of a Series of Functions Consider the series
$$F(x)=\sum_{n=1}^{\infty}\frac{\cos(nx)}{n}$$
with $0<a\le x\le b<2\pi$
Show that over $[a,b]$, $F$ is uniformly, but not absolutely, convergent.
 A: For uniform convergence use Dirichlet's criterio, as indicated by David's commnet.
For non absolute convergence:
$$
\sum\frac{|\cos(n\,x)|}{n}\ge\sum\frac{|\cos(n\,x)|^2}{n}=\sum\frac{1+\cos(2\,n\,x)}{2\,n}=\sum\frac{1}{2\,n}+\sum\frac{\cos(2\,n\,x)}{2\,n}.
$$
The first series is divergent, while the second is convergent. In fact, this shows that
$$
\sum_{n=1}^N\frac{|\cos(n\,x)|}{n}\ge\frac{\log N}{2}-C
$$
for some constant $C>0$.
A: It has already been observed that the series does not converge absolutely.
Next, we show that it converges uniformly in $[a,b]$.
First we find a uniform bound for the sequence:
$$
s_n(x)=\cos x+\cos 2x+\cdots+\cos nx=\mathrm{Re}\,\left(\mathrm{e}^{ix}+\mathrm{e}^{i2x}+\cdots+\mathrm{e}^{inx}\right)=\mathrm{Re}\,\frac{\mathrm{e}^{i(n+1)x}-\mathrm{e}^{ix}}{\mathrm{e}^{ix}-1},
$$
and
$$
\frac{\mathrm{e}^{i(n+1)x}-\mathrm{e}^{ix}}{\mathrm{e}^{ix}-1}=\frac{\mathrm{e}^{i(n+1/2)x}-\mathrm{e}^{ix/2}}{\mathrm{e}^{ix/2}-\mathrm{e}^{-ix/2}}=\frac{\mathrm{e}^{i(n+1/2)x}-\mathrm{e}^{ix/2}}{2i\sin(x/2)}
$$
and thus
$$
\lvert s_n(x)\rvert\le \frac{1}{\lvert\sin(x/2)\rvert}\le\max_{x\in [a,b]}\frac{1}{\lvert\sin(x/2)\rvert}=M.
$$
Note that $M<\infty$, since $0<a\le b<2\pi$.
Next we use the Abel's summation method
\begin{align}
\sum_{k=1}^n\frac{\cos nx}{n}&=\sum_{k=1}^n\frac{s_n(x)-s_{n-1}(x)}{n}=
\sum_{k=1}^n\frac{s_n(x)}{n}-\sum_{k=1}^n\frac{s_{n-1}(x)}{n} \\&=
\sum_{k=1}^n\frac{s_n(x)}{n}-\sum_{k=1}^{n-1}\frac{s_{n}(x)}{n+1}
=\frac{s_n(x)}{n}+\sum_{k=1}^{n-1}s_n(x)\left(\frac{1}{n}-\frac{1}{n+1}\right) \\
&=\sum_{k=1}^{n-1}\frac{s_n(x)}{n(n+1)}+\frac{s_n(x)}{n},
\end{align}
which shows that our series converges uniformly as 
$$
\left\vert\frac{s_n(x)}{n}\right\rvert\le \frac{M}{n}\to 0,
$$ 
and the series
$$
\sum_{k=1}^{n-1}\frac{s_n(x)}{n(n+1)},
$$ 
converges uniformly (and absolutely) due to comparison test since
$$
\left\vert\frac{s_n(x)}{n(n+1)}\right\rvert\le \frac{M}{n(n+1)} 
\quad\text{and}\quad \sum_{n=1}^\infty \frac{M}{n(n+1)}<\infty.
$$ 
A: Use the trigonometric identity
$$2\cos A\sin B=\sin(A+B)-\sin(A-B)$$
with, as indicated by DavidMitra, $A=nx$, $B=\frac12 x$. 
Then
$$2\sin(\tfrac12 x)\sum_{n=1}^N\frac{\cos(nx)}{n}=-\sin(\tfrac12 x)+\sum_{n=1}^{N-1}\frac{\sin((n+\tfrac12) x)}{n(n+1)}+\frac{\sin((N+\tfrac12) x)}{N}$$
where now uniform convergence of the middle sum follows from the classical majorant $\sum\frac1{n(n+1)}$.
