Theorem 1.(Abel’s test)Let be a decreasing sequence of positive real numbers such that and the series

converges for and diverges for . Then the power series converges for all .

*Proof. *Let with . For the functions and , take the principal branch. Then

Let , and denote the and partial sums by and , respectively. Then

Taking the modulus of both sides and applying the triangle inequality, along with our hypothesis that is decreasing, positive, we see that

Taking sufficiently large, we have that , for any given, so the partial sums converge by the Cauchy criterion.

Recall that a series is said to be **Abel-summable** if the limit

exists.

The following theorem says that every convergent series (in the underlying topology) is Abel-summable. A classic example of a series in which does not converge yet is Abel-summable is . Indeed, for , we have a geometric progression

as .

Theorem 2.(Abel’s Summation Theorem)Let be a sequence of complex numbers such that the series

converges for all . If the series converges, then

where lies in a

Stoltz anglefor some positive constant .

*Proof. *Substracting a constant if necessary, we may assume without loss of generality that . Fix a Stoltz angle . Setting and observing that , we can rewrite as

Let be given, and choose sufficiently large so that for all . By continuity, we can choose such that

For lying the Stoltz angle fixed above, we have the estimate

which completes the proof.

We remark that by scaling, Abel’s theorem generalizes to any series with radius of convergence .

Perhaps, the reader is still left with the question of why we care about Abel summability or what use is Abel’s theorem? First, the reader should note that given any bounded sequence of complex numbers , is the generating function associated to . Abel’s theorem gives a method to compute closed-form expressions for infinite series. Before we give an example, we recall the alternating series test from calculus.

Lemma 3.(Alternating Series Test)Let be a bounded, decreasing sequence of nonnegative numbers such that . Then the series

converges.

*Proof.* Fix . Denote the partial sum by . It is clear that . Observe that

It follows from the monotone convergence theorem and the squeeze theorem that and moreover,

Consider the sequence defined by , for . Then is convergent by the alternating series test, so

converges for . Since the geometric series convergence is uniform on a compact subsets of , we can compute , for , term-by-term to obtain

which implies that . Since as , we conclude from Abel’s theorem that

Great article! At the very end: “Since the geometric series converges is uniform…” might wanna correct to ‘convergence 😀

Thank you for both the kind words and the correction.