Hardy–Littlewood Tauberian theorem

Remarks

Karamata's proof shows first that $(1 - x) \sum_{n} a_n x^n P(x^n) \to \int_0^1 P(t) \, dt$ for every polynomial $P$, by checking it on monomials $P(t) = t^m$. Then it approximates the function $g$ with $g(t) = 1/t$ on $[e^{-1}, 1]$ and $g(t) = 0$ on $[0, e^{-1})$ from above and below by polynomials with close integrals. Positivity of $a_n$ makes these approximations compatible with the sums. Taking $x = e^{-1/n}$, one has $x^k g(x^k) = 1$ for $k \le n$ and $0$ for $k > n$, which isolates the partial sum $\sum_{k \le n} a_k$.

Example: for $a_n = 1 + (-1)^n$, one has $\sum a_n x^n = \frac{1}{1-x} + \frac{1}{1+x} \sim \frac{1}{1-x}$, and indeed $\sum_{k \le n} a_k$ is $n + 2$ or $n + 1$.

Positivity is needed. For $a_n = 1 + (-1)^n n$, one has $\sum a_n x^n = \frac{1}{1-x} - \frac{x}{(1+x)^2} \sim \frac{1}{1-x}$. But $\sum_{k=0}^{n} (-1)^k k$ is $\frac{n}{2}$ for even $n$ and $-\frac{n+1}{2}$ for odd $n$, so the partial sums $\sum_{k \le n} a_k$ are close to $\frac{3n}{2}$ or to $\frac{n}{2}$ and are not $\sim n$.

Going deeper

For a student

This theorem is the prototype of Tauberian theorems with positivity conditions, where no condition on the size of single terms is needed. In terms of $x = e^{-s}$ it says that $\sum a_n e^{-sn} \sim \frac1s$ as $s \to 0^+$ implies $\sum_{k \le n} a_k \sim n$; Karamata's Tauberian theorem extends it to Laplace–Stieltjes transforms with regularly varying growth. The same method of proof gives Littlewood's Tauberian theorem, and the theorem applies to counting problems in number theory where generating functions are easier to estimate than partial sums.