# 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

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.