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?

  • 2
    $\begingroup$ You sure that's the definition of the Cesaro sum? $\endgroup$ – Simply Beautiful Art May 11 '17 at 22:01
  • 2
    $\begingroup$ That's the definition of a Cesaro limit, not a Cesaro sum and he claims that the Cesaro limit doesn't exist. $\endgroup$ – spaceisdarkgreen May 11 '17 at 22:11
  • $\begingroup$ @SimplyBeautifulArt thanks for catching that! $\endgroup$ – man_in_green_shirt May 11 '17 at 22:17

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 $n$.

  • $\begingroup$ Sorry for the delay in accepting the answer, it took me a while to verify all the steps. Thanks for having put in the effort to write this! $\endgroup$ – man_in_green_shirt May 12 '17 at 21:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.