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The question is to find the $x$ such that series $\sum\limits_{n = 1}^\infty {\frac{{\cos nx}}{n}} $ converges. I can see if $x = k \pi$, then the series converges since it is alternating series. Also it converges when like $x = k\pi/2,x = k\pi/3,x = k\pi/4,...$ etc.

I have a solution that actually $x$ can be any real number except for $x=2k\pi,k=0,1,2,...$. Anyone can help explain why is this so?

Thanks!

Tuyet
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1 Answers1

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If $\alpha>1$, then the series converges absolutely. We assume from now on that $0<\alpha\le1$.

Theorem (Abel) If $(a_n)\in\mathbb{C}^\mathbb{N}$ and $(t_n)\in[0,+\infty[^\mathbb{N}$ are sequences such that :

  • $\exists M\ge0\,\forall n\in\mathbb{N},\,\left|\sum_{k=0}^na_k\right|\le M$

  • $\forall n\in\mathbb{N},\,t_{n+1}\le t_n$ and $\lim_{n\to\infty}t_n=0$

then the series $\sum_{n\ge0}t_na_n$ converges.

If you apply the above theorem to the special case $a_n=e^{inx}$ and $t_n=\frac 1{n^\alpha}$ (with $n$ starting at $1$), you will get that as soon as $x\in\mathbb{R}-2\pi\mathbb{Z}$, the series $\sum_{n\ge1}\frac{e^{inx}}{n^\alpha}$ converges and hence the same for its real part, that is $\sum_{n\ge1}\frac{\cos(nx)}{n^\alpha}$.

You need to show that, for $x\in\mathbb{R}-2\pi\mathbb{Z}$, the partial sums $\sum_{k=1}^ne^{ikx}$ define a bounded sequence, which is easy since :

$$\sum_{k=1}^ne^{ikx}=e^{ix}\sum_{k=0}^{n-1}\left(e^{ix}\right)^k=\frac{e^{ix}(1-e^{inx})}{1-e^{ix}}$$

and so :

$$\left|\sum_{k=1}^ne^{ikx}\right|\le\frac{2}{\left|1-e^{ix}\right|}$$

It should be added that, still for $x\in\mathbb{R}-2\pi\mathbb{Z}$, the series $\sum_{n\ge1}\frac{\cos(nx)}{n^\alpha}$ converges conditionally, which means that it does not converge absolutely. Here is short proof :

$$\left|\frac{\cos(nx)}{n^\alpha}\right|\ge\frac{\cos^2(nx)}{n^\alpha}=\frac{1+\cos(2nx)}{2n^\alpha}$$

But the series $\sum_{n\ge1}\frac{1}{2n^\alpha}$ diverges and the series $\sum_{n\ge1}\frac{\cos(2nx)}{2n^\alpha}$ converges (for the same reasons as for $\sum_{n\ge1}\frac{\cos(nx)}{n^\alpha}$) for all $x\in\mathbb{R}-\pi\mathbb{Z}$. It remains the possibility that $x=(2q+1)\pi$ : in that case, the conclusion is the same because $\sum_{n\ge1}\frac{\cos(nx)}{n^\alpha}$ becomes $\sum_{n\ge1}\frac{(-1)^n}{n^\alpha}$, which is a convergent alternating series.

Adren
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