Summation by Abelian means

For 0 =< λ_0 < λ_1 < λ_2 < ... with (λ_n) tending towards infinity and

f(t) = sum_{n = 0}^{infinity} (a_n).exp [(λ_n)t] converges for all positive real values of t,

If,

lim_{t->0} f(t) = s

then,

a_0 + a_1 + a_2 + ... = s , (A, λ)

Summation by Abelian means (A, λ) embodies a whole class of methods (which may indeed be further generalised, see Hardy, Divergent Series, pp71-73) depending on the sequence (λ_n) chosen, though each such method can be shown to be regular, linear and stable. We are interested in the simplest case where λ_n = n, whereupon we have with;

f(x) = sum_{n = 0}^{infinity} (a_n).x^n

where x = e^-t and hence convergent for |x| <1

if,

lim_{x -> 1^-} f(x) = s

(this notation implies the left hand limit)

then,

a_0 + a_1 + a_2 + ... = s , (A)

The 'A' stands in for Abel summation, which one might find a curious choice of name given Abel's attitude towards divergent series. Indeed, the present method is more fitting for the legacy of Euler or Poisson than Abel (we shall later see the intimate connection between a more powerful summation method attributed to Euler and Abel summation). However, it was Abel who employed a similar argument (under different conditions to do with limiting partial sums of course) for a theorem to do with Cauchy summation and the proper articulation has associated the method to his name.

The method is less powerful than the (B) and (B') methods discussed in the previous post, failing even to sum the geometric series outside its radius of convergence. However, it possesses the stability that Borel's methods in general do not and also generalise the Cesaro summation which we shall discuss in the next post.

Note: We may also mention in passing, the method of Lambert summation which also relies on a trick in the limit as with Borel's integral method but in a form more familiar with Abel summation.

We say,

a_0 + a_1 + a_2 + ... = s , (L)

if,

lim_{x -> 1^-} (1 - x). [sum_{n = 1}^{infinity} n.(a_n).(r^n) / (1 - x^n)] = s

Evaluating the limit (1 - x) / (1 - x^n) at each n shows how this method works.

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