Hardy's Tauberian theorem for Cesàro means

Remarks

Let $s_n$ be the partial sums and $\sigma_n$ their Cesàro means. For $m > n$ one has the identity $$s_n = \frac{(m+1)\sigma_m - (n+1)\sigma_n}{m - n} - \frac{1}{m - n} \sum_{k=n+1}^{m} (s_k - s_n).$$ The first term tends to $s$ when $m \approx (1 + \varepsilon) n$ and $n \to \infty$. The second term is at most $\max_{n < k \le m} \abs{s_k - s_n} \le K \frac{m - n}{n} \approx K\varepsilon$, so $\limsup \abs{s_n - s} \le K\varepsilon$ for every $\varepsilon > 0$.

The Tauberian condition cannot be dropped. The series $1 - 1 + 1 - \cdots$ is Cesàro summable to $\frac12$ and diverges; here $n \abs{a_n} = n$ is unbounded. For real $a_n$, the one-sided condition $n a_n \ge -K$ is already enough.

Going deeper

For a student

This is one of the first Tauberian theorems in which the condition $n a_n \to 0$ of Tauber's theorem is relaxed to $n a_n = O(1)$. The analogous result for Abel summation is Littlewood's Tauberian theorem, which is harder, because Abel summation is stronger than Cesàro summation. Hardy's theorem is used to show that a Fourier series of a function of bounded variation converges wherever its Cesàro means do, since its coefficients are $O(1/n)$.