Prove that $a_n:=\sin(\log(n))$ isn't Cesaro summable

Question

The Cesaro limit is defined to be $$\lim\limits_{N\to\infty}\frac{\sum\limits_{n=1}^Na_n}{N}.$$

On his latest blog post, Terence Tao mentions that this sequence isn't Cesaro summable.

How does one go about proving such a result?

Answer 1

First, we need some high order mean value theorem.

For any $C^2$ function $f$ on $[a,b]$ and $x\in (a,b)$, there is a $c_x\in (a,b)$ such that $$f(x) = f(a) + \frac{f(b)-f(a)}{b-a}(x-a) -\frac12f''(c_x)(x-a)(b-x)$$ If $M$ is an upper bound of $|f''(x)|$ on $[a,b]$, integrating above expression over $[a,b]$ leads to a bound of the form:

$$\left|\int_a^b f(x) dx - \frac12(f(a)+f(b))(b-a)\right| \le \frac{M}{12}(b-a)^3$$

Apply this to $\sin(\log x)$ over $[k,k+1]$ for positive integer $k$ and notice $$|\sin(\log x)''| = \left|\frac{\sin(\log x) + \cos(\log x)}{x^2}\right| \le \frac{\sqrt{2}}{x^2} \le \frac{\sqrt{2}}{k^2}\quad\text{ for } x \in [k,k+1]$$ We obtain $$\left|\int_k^{k+1}\sin(\log x) dx - \frac12\bigg(\sin(\log k)+\sin(\log(k+1))\bigg)\right|\le \frac{\sqrt{2}}{12 k^2}$$ Summing this from $k = 1$ to $n-1$, we obtain an estimation of the partial sums $$\begin{align} \sum_{k=1}^n a_k = \sum_{k=1}^n\sin(\log k) &= \frac12\sin(\log n) + \int_1^n\sin(\log x)dx + \epsilon_n\\ &= \frac12\left[ (n+1)\sin(\log n) - n\cos(\log n) + 1\right] + \epsilon_n \end{align} $$ and the error term $\epsilon_n$ is a bounded sequence. $$| \epsilon_n |\le \frac{\sqrt{2}}{12}\sum_{k=1}^{n-1}\frac{1}{k^2} \le \frac{\sqrt{2}}{12}\zeta(2) = \frac{\sqrt{2}\pi^2}{72} < \infty$$

Using this, we can conclude

$$\lim_{n\to\infty}\left(\frac1n \sum_{k=1}^n a_k - \frac12(\sin\log(n) - \cos\log(n))\right) = 0$$

This means the Cesaro limit $\displaystyle\;\lim_{n\to\infty} \frac1n \sum_{k=1}^n a_k\;$ doesn't exist at all.
Instead, $\displaystyle\;\frac1n \sum_{k=1}^n a_k\;$ oscillates like $\frac12(\sin(\log n) - \cos(\log n))$ for large $n$.