$\sum_{n=1}^\infty a_n \cos nx$ unbounded near $0$ if $\sum a_n$ diverges?

If $$a_n$$ is a decreasing positive sequence and tends to $$0$$, and given$$\sum_{n=1}^\infty a_n=+\infty$$
can we prove that $$\lim_{x\rightarrow 0}\sum_{n=1}^{\infty} a_n \cos nx =+\infty$$ or at least prove the series above is unbounded for $$x$$ in a neighborhood of $$0$$?
For the power series $$\sum_{n=1}^\infty a_n \left(1-x\right)^n$$ the conclusion holds, because $$\left(1-x\right)^n$$ is positive.
I wonder if trigonometric series can have the similar conclusion, so I tested several $$a_n$$, plotted the graph, and discovered that it is probably true.
However, since $$\cos nx$$ is not identically positive, it is hard to give a rigorous proof, and I cannot give a counter-example either.
Anyone has some ideas?

Now I can answer half of the question.
The conclusion is as below:

Theorem 1. For monotone decreasing sequence $$a_n$$ satisfying $$a_n\rightarrow 0$$ and $$\sum a_n$$ diverges, the sum $$\sum_{n=1}^\infty a_n\cos nx$$ must be UNBOUNDED in a neighborhood of $$0$$.

The proof goes like this:

First we have $$a_n\ge 0$$. By Dirichlet’s method, the series $$S\left(x\right)=\sum_{n=1}^\infty a_n\cos nx$$ converges uniformly on $$\left(\delta,\pi\right]$$ for any $$\delta>0$$, therefore is continuous on it.
If $$S$$ is not integrable on $$\left[0,\pi\right]$$, it must be unbounded on $$\left(0,\delta\right)$$ for some $$\delta>0$$.
If $$S$$ is integrable on $$\left[0,\pi\right]$$, then the Fourier series of $$S$$ is $$S\left(x\right)$$ itself (See Hardy Fourier Series §3.10 Thm.46).
By the properties of Fejér kernel, we can write the n-th Cesàro partial sum of $$S$$ at $$x=0$$ as $$\sigma_n=\frac{1}{\pi}\int_0^\pi S\left(t\right)F_n\left(t\right)dt$$
where $$F_n\ge 0$$ is the Fejér kernel.
Suppose $$\left|S\left(x\right)\right|\le M$$, then $$\sigma_n\le \frac{M}{\pi}\int_0^\pi F_n\left(t\right)dt=M$$ As the series $$\sum a_n$$ is positive and divergent, the n-th Cesàro partial sum $$\sigma_n\rightarrow+\infty$$ as $$n\rightarrow \infty$$(by Stolz’s theorem), which leads to contradiction.
In conclusion, $$S$$ must be unbounded near $$0$$. $$\blacksquare$$

But it remains to prove or disprove $$S\left(x\right)\rightarrow+\infty$$ as $$x\rightarrow0$$. It still needs tougher work.

There are also two little theorem about divergence to $$+\infty$$ of $$S\left(x\right)$$:

Theorem 2. If $$a_n$$ is a CONVEX sequence(i.e.$$a_n+a_{n+2}\ge 2a_{n+1}\,\forall n$$) tends to $$0$$, and $$\sum a_n$$ diverges, then $$S\left(x\right)\rightarrow +\infty$$ when $$x\rightarrow 0$$.

Theorem 3. If $$a_n$$ is defined as Thm.1, and $$S$$ is NOT integrable on $$\left[0,\pi\right]$$, $$S$$ must have infinitely many zeroes near $$0$$, hence does not tend to $$+\infty$$.

Proof(Thm.2):

Using summation by parts twice and we can get $$S\left(x\right)=\sum_{n=1}^\infty \left(a_n+a_{n+2}-2a_{n+1}\right)\frac{1-\cos\left(n+1\right)x}{4\sin^2 \frac{x}{2}}$$ which is a positive-term series.
There exists a constant $$C>0$$, for $$0, $$\frac{1-\cos nx}{4\sin^2 \frac{x}{2}}\ge Cn^2$$ Therefore $$S\left(x\right)\ge C\sum_{n=1}^{\left[\frac{1}{x}\right]-1} \left(n+1\right)^2 \left(a_n+a_{n+2}-2a_{n+1}\right)$$ Then we need to prove that RHS is divergent.
We need a lemma: If $$b_n$$ is decreasing and tends to $$0$$, $$\sum n^k b_n$$ diverges, then $$\sum n^{k+1}\left(b_n-b_{n+1}\right)$$ also diverges.
Since $$\Delta n^{k+1}=n^{k+1}-\left(n-1\right)^{k+1}\sim \left(k+1\right) n^k$$ we know that $$\sum \Delta n^{k+1}\,b_n$$ diverges.
For $$M>0$$, There exists $$N_0$$ s.t. $$\sum_{n=1}^{N_0}\Delta n^{k+1}\,b_n>M+1$$ As $$b_n\rightarrow 0$$, there exists $$N_1$$, for all $$N>N_1$$ we have $$N_0^{k+1}b_N<1$$.
Then for $$N>\max\left\{N_0,N_1\right\}$$,\begin{aligned}\sum_{n=1}^N n^{k+1}\left(b_n-b_{n+1}\right)&=\sum_{n=1}^N \Delta n^{k+1}\left(b_n-b_{N+1}\right)\\&\ge\sum_{n=1}^{N_0}\Delta n^{k+1} \left(b_n-b_{N+1}\right)\\&=\sum_{n=1}^{N_0} \Delta n^{k+1}\,b_n -N_0^{k+1} b_{N+1}\\&>M+1-1=M\end{aligned} Hence by the lemma we can get $$\sum a_n =+\infty \Rightarrow \sum n \left(a_n-a_{n+1}\right) =+\infty \Rightarrow \sum n^2 \left(a_n+a_{n+2}-2a_{n+1}\right)=+\infty$$ which finally proves the theorem. $$\blacksquare$$

The proof of Thm.3 needs two lemmas as follow.

Lemma 1. If $$b_n$$ is decreasing and $$nb_n\rightarrow 0$$, then $$\sum_{n=1}^\infty b_n \sin nx$$ converges uniformly on $$\mathbb{R}$$.
Lemma 2. $$S$$ has an antiderivative $$\sum _{n=1}^\infty \frac{a_n}{n}\sin nx$$ and $$\lim_{\delta\rightarrow 0^+}\int _\delta ^\pi 2S\left(x\right) \cos mx dx=\pi a_m$$

The proof of Lemma.1 is a little complex but not difficult, and is taken from an exercise book in mathematical analysis. We will just assume it right.

Proof of Lemma.2:

Notice that the series of $$S$$ converges uniformly in $$\left[\delta,\pi\right]$$, hence we can integrate by terms \begin{aligned}\int_\delta^\pi 2S\left(x\right)\cos mx\,dx&=\sum_{n=1}^\infty \int_\delta^\pi 2a_n \cos nx \cos mx\,dx\\&=\sum_{n=1}^\infty \frac{a_n}{n+m}\sin\left(n+m\right)\delta +\sum_{n\ne m}\frac{a_n}{n-m}\sin \left(n-m\right)\delta +a_m\left(\pi-\delta+\frac{\sin 2m\delta}{2m}\right)\\&\overset{def}{=}\Theta_1\left(\delta\right)+\Theta_2\left(\delta\right)+a_m\left(\pi-\delta+\frac{\sin 2m\delta}{2m}\right)\end{aligned} Clearly by lemma 1, $$\Theta_1$$ and $$\Theta _2$$ both converge uniformly on $$\mathbb{R}$$, hence continuous at 0.
So we have $$\lim_{\delta\rightarrow 0^+}\int_\delta^\pi 2S\left(x\right)\cos mx\,dx=\pi a_m$$
Letting $$m=0$$ we can get that $$\int_x^\pi S\left(t\right) dt=\sum_{n=1}^\infty \frac{a_n}{n}\sin nx$$ By the continuity of $$S$$ we finish the proof. $$\blacksquare$$

Back to the proof of Thm.3

Note that $$S$$ is not integrable, that means $$\int_0^\pi \left|S\left(t\right)\right| dt= +\infty$$ But the improper integral of $$S$$ converges to $$0$$ as discussed above.
That means S must have infinitely many positive parts and negative parts near $$0$$, by continuity there must be infinitely many zeroes. $$\blacksquare$$

So if we are going to find $$S$$ such that $$\lim_{x\rightarrow 0}S\left(x\right)\ne +\infty$$, we can just find an unintegrable $$S$$, and the coefficient $$a_n$$ must not be convex.

• Would $S’(x)$ give some useful information? – Szeto Jan 28 at 23:20
• @Szeto But $S$ is not necessarily differentiable, even nowhere differentiable. – Antimonius Jan 29 at 3:48

All I can think of at this moment is the following very restricted partial result:

Claim. If $$(a_n)$$ is non-negative, non-increasing and $$n a_n \leq C$$ for some constant $$C > 0$$, then

$$\lim_{n\to\infty} \sum_{n=0}^{\infty} a_n \cos(n x) = \sum_{n=0}^{\infty} a_n,$$

regradless of the convergence of $$\sum_{n=0}^{\infty} a_n$$.

Since the claim is obvious if $$\sum_{n=0}^{\infty} a_n < \infty$$, we may focus on the case $$\sum_{n=0}^{\infty} a_n = \infty$$. To this end, define an auxiliary sequence $$(b_n)$$ by $$a_n - a_{n+1} = b_n$$. Then

• $$b_n \geq 0$$ and $$a_m = a_n + \sum_{k=m}^{n-1} b_k$$ for $$0 \leq m < n$$,
• $$a_n = \sum_{k=n}^{\infty} b_k$$, and
• $$\sum_{k=0}^{\infty} (k+1)b_k = \sum_{n=0}^{\infty} a_n = \infty$$. (This follows from Tonelli's theorem.)

Moreover, we can apply summation by parts

\begin{align*} \sum_{n=0}^{N} a_n \cos(nx) &= a_N \sum_{n=0}^{N} \cos(nx) + \sum_{n=0}^{N-1} \left( \sum_{k=n}^{N-1} b_k \right) \cos(nx) \\ &= a_N D_N(x) + \sum_{k=0}^{N-1} b_k D_k(x), \end{align*}

where $$D_k(x) = \sum_{n=0}^{k} \cos(nx)$$. Now if $$x \in (0, \pi]$$, then $$D_n(x)$$ is bounded in $$n$$. So, as $$N\to\infty$$, the above converges to

\begin{align*} \sum_{n=0}^{\infty} a_n \cos(nx) = \sum_{k=0}^{\infty} b_k D_k(x). \end{align*}

Now notice that $$D_k(x) = \cos(kx/2)\sin((k+1)x/2)/\sin(x/2)$$. So, if $$x \in (0, 1)$$ and $$k \leq 1/x$$, then

$$D_k(x) \geq \frac{\cos(1/2) \cdot \frac{2}{\pi} (k+1)x/2}{x/2} = c(k+1)$$

for $$c = \frac{2}{\pi}\cos(1/2) > 0$$. So

\begin{align*} \sum_{n=0}^{\infty} a_n \cos(nx) &\geq \sum_{0 \leq k \leq 1/x} b_k D_k(x) - \sum_{k > 1/x} \frac{b_k}{\sin(x/2)} \\ &\geq c \sum_{0 \leq k \leq 1/x} (k+1)b_k - \frac{a_{\lceil 1/x \rceil}}{\sin(x/2)}. \end{align*}

Since $$a_{\lceil 1/x \rceil} = \mathcal{O}(x)$$ as $$x \to 0^+$$, we have $$\frac{a_{\lceil 1/x \rceil}}{\sin(x/2)} = \mathcal{O}(1)$$, and so, letting $$x\to 0^+$$ proves the desired claim.

• Nice answer. It can also explain an interesting phenomenon that $S\left(x\right)$ diverges to $\infty$ at the same rate of $\sum_{n=1}^{\left[\frac{1}{x}\right]}a_n$ under certain condition. And recently I proved some other little theorems of the divergence to $\infty$ of $S\left(x\right)$. – Antimonius Feb 1 at 3:18