Tauber's theorem

Remarks

The idea of the proof is to compare the partial sum $s_N = \sum_{n \le N} a_n$ with $f(x)$ for $x = 1 - 1/N$. The difference is at most $\sum_{n \le N} \abs{a_n} (1 - x^n) + \sum_{n > N} \abs{a_n} x^n$. Use $1 - x^n \le n(1 - x)$ in the first sum, which is then at most the average of $n \abs{a_n}$ over $n \le N$, and this tends to $0$. In the second sum, $\abs{a_n} \le \varepsilon / n \le \varepsilon / N$ for large $N$, so it is at most $\varepsilon / (N(1 - x)) = \varepsilon$.

The theorem is a partial converse of Abel's theorem, which gives $\lim_{x \to 1^-} f(x) = \sum a_n$ whenever the series converges. The condition on $n a_n$ cannot be dropped. For $a_n = (-1)^n$, one has $f(x) = 1/(1 + x) \to 1/2$, but $\sum (-1)^n$ diverges, and $n a_n$ does not tend to $0$. Littlewood showed that the weaker condition $n a_n = O(1)$ is enough; see big O notation.

Going deeper

For a student

This is the first Tauberian theorem, and it gave its name to the whole family of Abelian and Tauberian theorems. An Abelian theorem passes from ordinary convergence to convergence of a summation method; a Tauberian theorem goes back, under an extra size condition on the terms. Later results of this type, such as the Hardy–Littlewood and Wiener Tauberian theorems, are central tools in analytic number theory.