0
$\begingroup$

Is anything known about this kind of `infinite' Cesàro summation (or any related types of summation)?

If we have a function we wish to sum $f(n)$, but $$ S^0[f] = \sum_{n=1}^\infty f(n) $$ diverges, so we take the average of partial terms (Cesàro summation) $$ s_1=\frac{1}{n_1}\sum_{n_0=1}^{n_1} f(n_0)\\ S^1[f] = \lim_{n_1 \to \infty}s_1=\lim_{n_1 \to \infty}\frac{1}{n_1}\sum_{n_0=1}^{n_1} f(n_0) $$ but say this also diverges, so we iterate $$ s_2 = \frac{1}{n_2}\sum_{n_1=1}^{n_2} \frac{1}{n_1}\sum_{n_0=1}^{n_1} f(n_0)\\ S^2[f] = \lim_{n_2 \to \infty} s_2 = \lim_{n_2 \to \infty}\frac{1}{n_2}\sum_{n_1=1}^{n_2} \frac{1}{n_1}\sum_{n_0=1}^{n_1} f(n_0) $$ then what can be said about $S^{\infty}$, if this is performed through a limit on the partial $s_k$'s?

Some examples:

For, $f(n)=n$ the intermediate terms look like $$ s_k=\frac{(2^k-1)n_k + n_k^2}{2^k n_k}\\ S^{\infty}[f] = \lim_{k \to \infty} \frac{(2^k-1)n_k + n_k^2}{2^k n_k} = 1 $$

For, $f(n)=n^2$ the intermediate terms appear to look like $$ s_k=\frac{(2^{k+1}-3^{k+1}+6^k)n_k + 3(3^k-2^k)n_k^2 + 2^k n_k^3}{6^k n_k}\\ S^{\infty}[f] = \lim_{k \to \infty} \frac{(2^{k+1}-3^{k+1}+6^k)n_k + 3(3^k-2^k)n_k^2 + 2^k n_k^3}{6^k n_k} = 1 $$

  • For $f(n)=n+1$, $S^{\infty}[f]=2$
  • For $f(n)=n-1$, $S^{\infty}[f]=0$
  • It seems that for $f(n)=n+k$, $S^{\infty}[f]=1+k$

The most interesting case seems to be $f(n)=\frac{1}{n}$ which then relates to $\zeta(1)$. We have

\begin{equation} s_1 = \frac{H(n)}{n} \\ s_2 = \frac{H(n)^2+H^{(2)}(n)}{2n} \\ s_3 = \frac{H(n)^3+3H(n)H^{(2)}(n)+2H^{(3)}(n)}{6n} \\ s_4 = \frac{H(n)^4+6H(n)^2H^{(2)}(n)+3H^{(2)}(n)^2 + 8 H(n)H^{(3)}(n)+6H^{(4)}(n)}{24n} \\ \end{equation} where the sum in $s_j$ seems to be over products of harmonic numbers $H^{(k_i)}(n)$ such that $\sum_i k_i = j$. It is not clear how to take the limit for $s_\infty$. The coefficients seem related to A102189 (and A036039), which makes sense. The paper referenced in A102189 does mention similar coefficients in a related expansion for multiple zeta values but in terms of powers of $\zeta(k)$ rather than powers of $H^{(k)}(n)$. In some sense the infinite limit $s_{\infty}$ is the related to $\zeta(1,1,1,\cdots)$ for an infinite number of $1$'s.

This seems to relate to this Wikipedia article. Which would imply $$ s_k(n) = \frac{1}{n}\sum_{m_1 + 2m_2 + \cdots + km_k = k \atop m_1\ge 0, \ldots, m_k\ge 0} \prod_{i=1}^k \frac{H^{(i)}(n)^{m_i}}{m_i ! i^{m_i}} $$ but it is not clear how an infinite limit $k \to \infty$ would be taken.

Edit:

Based on an identity here, it would seem that $$ \text{Li}_k(z) = \sum_{n \ge 1}\frac{(-1)^{n-1}n s_k(n)z^n}{(1-z)^{n+1}} $$

$\endgroup$
  • 1
    $\begingroup$ Your definition of $s_2$ should likely have $\sum_{n_1=1}^{n_2}$ rather than $\sum_{n_1=1}^{\infty}$. $\endgroup$ – Michael Jul 4 '18 at 13:28
  • $\begingroup$ @Michael You are right, thank you! $\endgroup$ – Benedict W. J. Irwin Jul 4 '18 at 13:29
  • 1
    $\begingroup$ Your $S^{\infty}$ limit also is not clear since we don't know how fast $n_k$ increases in relation to $2^k$ (for the $f(n)=n$ case). How do you evaluate the limit to $1$? $\endgroup$ – Michael Jul 4 '18 at 13:32
  • $\begingroup$ @Michael Sorry that is just a notational problem, $n_k$ is just the variable form the $k^{th}$ iteration. It is not a function of $k$. It will always be a summation index after any number of iterations. $\endgroup$ – Benedict W. J. Irwin Jul 4 '18 at 13:38

Your Answer

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

Browse other questions tagged or ask your own question.