Does $\sum_{n=1}^\infty\frac{\cos\left(\left(2n-1\right)x\right)}{2n-1}$ converge absolutely $$\sum_{n=1}^\infty\frac{\cos\left(\left(2n-1\right)x\right)}{2n-1}$$
We have that $\sum_{n=1}^\infty|\frac{\cos\left(\left(2n-1\right)x\right)}{2n-1}| \leq \sum_{n=1}^\infty \frac{1}{2n-1}$. However the series on the right diverges as $\sum_{n=1}^\infty \frac{1}{n}$. Which convergence test is most suitable for this series?
 A: Multiply with $2\sin(x)$ and use
$$
2\sin x\cos((2n-1)x)=\sin(2nx)-\sin(2(n-1)x)
$$
to get
$$
2\sin x·\sum_{n=1}^N\frac{\cos((2n-1)x)}{2n-1}=\frac{\sin(2Nx)}{2N-1}+2\sum_{n=1}^{N-1}\frac{\sin(2nx)}{4n^2-1}
$$
where you get absolute and uniform convergence on the right side so that the only singularity results from the division by $\sin x$.
A: Dirichlet's Test for convergence states that if the partial sums of $\sum_{n=1}^\infty x_n$ are bounded, and if $(y_n)$ is a sequence satisfying $y_1\geq y_2\geq\cdots\geq 0$ with $\lim y_n=0$, then the series $\sum_{n=1}^\infty x_ny_n$ converges (from Abbott's Understanding Analysis).
In your case $y_n=\frac{1}{2n-1}$ clearly satisfies the requirements. It remains to be shown that taking $x_n=\cos((2n-1)x)$, one has that the partial sums of $\sum_{n=1}^\infty x_n$ are bounded.
It can be shown inductively that $$\sin\frac{x}{2}\cdot\left(\cos x+\cos2x+\cdots+\cos nx\right)=\cos\frac{(n+1)x}{2}\cdot\sin\frac{nx}{2}.$$
If $x\neq 2k\pi$ for any $k\in\mathbb{Z}$, then $$\left|\cos x+\cos2x+\cdots+\cos nx\right|=\frac{\left|\cos\frac{(n+1)x}{2}\cdot\sin\frac{nx}{2}\right|}{\left|\sin\frac{x}{2}\right|}\leq\frac{1}{\left|\sin\frac{x}{2}\right|}.$$
Thus, the partial sums of $\sum_{n=1}^\infty\cos(nx)$ are bounded as long as $x\neq 2k\pi$ for any $k\in\mathbb{Z}$. But also, the partial sums of $\sum_{n=1}^\infty\cos(2nx)$ are bounded provided $x\neq k\pi$ for any $k\in\mathbb{Z}$ (replace $x$ above with $2x$). The partial sums of $\sum_{n=1}^\infty x_n$ can all be written as a difference between the partial sums of these two bounded series, and are therefore bounded. This justifies the use of Dirichlet's Test, and the series converges for all $x\neq k\pi$ for $k\in\mathbb{Z}$.
A: If $0<x<\pi,$ then
$$\sum_{n=1}^{\infty}\frac{|\cos(2n-1)x\,|}{2n-1}=\infty.$$
Proof: The above is
$$\tag 1 \ge \sum_{n=1}^{\infty}\frac{\cos^2 (2n-1)x}{2n-1} = \sum_{n=1}^{\infty}\frac{1+\cos (4n-2)x}{4n-2} .$$
Now $\sum 1/(4n-2) = \infty,$ while $\sum (\cos (4n-2)x)/(4n-2)$ converges by Dirichlet's test. Thus the series in $(1)$ diverges.
