# An elucidated proof of Tauber's theorem

Titchmarsh’s (in)famous The Theory of Functions proves Tauber’s theorem in five lines. It’s a wonderful proof, but if you’re like me and prefer physical explanations of proofs—or just take awhile to wrap your head around analytical proofs in general—this concision can be a little overwhelming. Here, I’ll proove Tauber’s theorem in essentially the same manner that Titchmarsh does, but I’ll explain the steps in much greater detail.

Tauber’s theorem states the following: if $$a_n \in o(1/n)$$ and the function $$f(x) \rightarrow s$$ as $$x \rightarrow 1$$, then the series $$\sum_{n=0}^{\infty} a_n \rightarrow s$$. The applied appeal of this theorem is clear: if we know that a given series $$\sum_{n=0}^{\infty} a_n$$ converges, then we know the power series representation of the function $$f$$ is convergent on the open unit disk.

First, a note on notation. Writing $$g(n) \in o(h(n))$$ means that $$\frac{g(n)}{h(n)} \rightarrow 0$$ as $$n$$ approaches infinity. So, here we’re saying that $$na_n \rightarrow 0$$ as n gets large. Representing $$f(x) = \sum_{n=0}^{\infty} a_nx^n$$, a straighforward way to prove the theorem means showing that

as $$x \rightarrow 1$$. We’ll find it convenient to rewrite the above statement as

$\sum_{n=N+1}^{\infty} a_nx^n - \sum_{n=0}^{N} a_n(1-x^n) \rightarrow 0$ as $$x \rightarrow 1$$, setting $$N = \frac{1}{1-x}$$. (If you have trouble seeing the equivalence, expand the last sum and compare terms.) The strategy now is to show that the two sums above are each less than $$\epsilon$$ for some value of $$N$$. Because of the condition on the coefficients $$a_n$$, we can safely say that $$na_n < \epsilon$$ for all $$n > n_0$$. We set $$n_0 = N$$ and write the first sum as

as $$\frac{\epsilon}{N+1} > a_n$$ for all $$n > N+1$$. Now, from the formula for the geometric series $$\sum_{n=0}^{\infty}x^n = \frac{1}{1-x}$$ for $$x$$ in the open unit disk, we can write a new upper bound:

Expanding the denominator and recalling the bounds on $$x$$, we thus have

which implies that $$\sum_{n=N+1}^{\infty} a_nx^n < \epsilon$$.

Now, considering the second sum $$\sum_{n=0}^{N} a_n(1-x^n)$$, we recall that $$1-x^n$$ can actually be represented as the finite telescoping sum

Clearly, $$(1-x)\sum_{j=0}^{n-1} x^j < n(1-x)$$, so the upper bound on the second sum is

which we can rewrite as $$\frac{1}{N}\sum_{n=0}^{N} na_n \rightarrow 0$$ as $$x \rightarrow 1$$. Thus, the sum of the absolute values of both sums is less than $$2\epsilon$$, and the proof is completed.

Written on July 29, 2016