Wikipedia:Reference desk/Archives/Mathematics/2019 December 21

= December 21 =

Convergence of Cesàro means: how would a nonstandard proof look?
It is a rather well-known result in basic analysis that if we have a real sequence $$(a_n)$$ such that $$a_n \rarr l$$ for some real limit $$l$$, and we define $$b_n = \frac{1}{n} \sum ^{n} _{k=1} a_k$$, then $$b_n \rarr l$$ as well. The standard proof (which I know) proceeds along these lines: given $$\epsilon > 0$$, pick some $$N$$ such that for all $$n \geq N$$, $$|a_k - l| < \epsilon$$. Then split the sum for $$|b_n - l| = \left|\frac{1}{n} \sum ^{n} _{k=1} (a_k - l)\right|$$ to bits before and after term $$N$$ using the triangle inequality, and write $$A$$ for the maximum of the first $$N$$ of the $$|a_k - l|$$. The sum of the terms after term $$N$$ is bounded by $$\frac {n - N}{n} \cdot \epsilon < \epsilon$$, and the sum of the terms before and including it is bounded by $$\frac{1}{n} \cdot N \cdot A$$, which is less than $$\epsilon$$ if we pick $$n$$ large enough. The result follows immediately.

My question is, how do you prove this result in nonstandard analysis with the infinitesimal-based definition of a limit? I am tempted to mimic the above: let $$H$$ be an infinite hyperinteger and consider $$b_H - l = \frac{1}{H} \sum ^{H} _{i=1} (a_i - l)$$. If $$i$$ is finite then the respective term is a finite number over an infinite one, therefore infinitesimal. And if $$i$$ is infinite then it is an infinitesimal over an infinity, so a second-order infinitesimal. The difficulty is that it seems to me that in the first half we are summing infinitely many infinitesimals, so we don't get the bound that we should have (as we need to prove that the entire sum is infinitesimal). How should I proceed? (I should mention that I am not really very experienced with nonstandard analysis, so I may be just misunderstanding something simple...) Double sharp (talk) 04:04, 21 December 2019 (UTC)
 * This seems like a good question that deserves a response. If you don't get an answer here then you might try mathoverflow; they seem to have a few people with expertise in nonstandard analysis. --RDBury (talk) 16:36, 24 December 2019 (UTC)