# 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?

• Seems tough because at $x=0$ the sum diverges, while at $x=\pi/2$ it converges. Is this from introductory calculus or more advanced analysis? – I. Cavey Jan 20 '16 at 18:18
• @I.Cavey The sum above is actually the Fourier series of $f\left(x\right)=\ln\sqrt{\cot\frac{x}{2}}$ and I am also trying to examine whether it's uniformly convergent and its convergence in the mean. – Emir Šemšić Jan 20 '16 at 18:23
• That Fourier series can't be uniformly convergent on $[0,2\pi]$ because it converges to an unbounded function. – zhw. Jan 20 '16 at 19:42
• @zhw. Weird, that means that integrating term by term can't be done. But I am asked to find the integral of the above series(using the series representation) – Emir Šemšić Jan 20 '16 at 19:48
• What interval are we working on by the way? Seems to me there is a problem with $\sqrt {\cot (x/2)}.$ – zhw. Jan 20 '16 at 21:00

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 $k\in\mathbb{Z}$.

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 $\sin x$.

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.