Frobenius theorem on Cesàro and Abel summability

Remarks

The sequence $(\sigma_n)$ converges, so it is bounded, and then $s_n = (n+1)\sigma_n - n\sigma_{n-1}$ and $a_n = s_n - s_{n-1}$ are $O(n)$; hence the power series converges for $\abs{x} < 1$. Multiplying by $\frac{1}{1-x}$ twice gives, for $\abs{x} < 1$, $$f(x) = (1 - x)^2 \sum_{n \ge 0} (n + 1) \sigma_n x^n.$$ Since $(1 - x)^2 \sum_{n \ge 0} (n+1) x^n = 1$, one gets $f(x) - s = (1 - x)^2 \sum_{n \ge 0} (n + 1)(\sigma_n - s) x^n$. Splitting the sum at a large $N$ where $\abs{\sigma_n - s} < \varepsilon$ for $n > N$ shows that this tends to $0$ as $x \to 1^-$.

Example: $1 - 1 + 1 - \cdots$ is Cesàro summable to $\frac12$, and indeed $\sum (-1)^n x^n = \frac{1}{1+x} \to \frac12$.

The converse fails: $1 - 2 + 3 - 4 + \cdots$ is Abel summable to $\frac14$, since $\sum (-1)^n (n+1) x^n = \frac{1}{(1+x)^2}$, but it is not Cesàro summable of order $1$.

Going deeper

For a student

This is an Abelian theorem in the sense of summability: it compares two methods and shows an inclusion with consistency. It is a strengthening of Abel's theorem, because every convergent series is Cesàro summable to its sum. The same argument shows that Cesàro summability of any order $k$ implies Abel summability to the same value.