Abel's theorem

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Template:More footnotes In mathematics, Abel's theorem for power series relates a limit of a power series to the sum of its coefficients. It is named after Norwegian mathematician Niels Henrik Abel, who proved it in 1826.[1]

Theorem

Let the Taylor series G(x)=∑k=0∞akxk be a power series with real coefficients ak. Suppose that the series ∑k=0∞ak converges. Then G(x) is continuous from the left at x=1, that is, limx→1−G(x)=∑k=0∞ak.

The same theorem holds for complex power series G(z)=∑k=0∞akzk, provided that z→1 entirely within a single Stolz sector, that is, a region of the open unit disk where |1−z|≤M(1−|z|) for some fixed finite M>1. Without this restriction, the limit may fail to exist: for example, the power series ∑n>0z3n−z2⋅3nn converges to 0 at z=1, but is unbounded near any other point of the form eπi/3n, so the value at z=1 is not the limit as z tends to 1 in the whole open disk.

The convergence of ∑k=0∞ak implies that the radius of convergence of the power series ∑k=0∞akzk is at least 1, ensuring convergence for |z|<1.

Also, by the uniform limit theorem, G(z) is continuous on the real closed interval [0,t] for t<1, by virtue of the uniform convergence of the series on compact subsets of the disk of convergence (by the Weierstrass M-test). Abel's theorem allows us to say more, namely that the restriction of G(z) to [0,1] is continuous.

Stolz sector

20 Stolz sectors, for M ranging from 1.01 to 10. The red lines are the tangents to the cone at the right end.

The Stolz sector |1−z|≤M(1−|z|) has explicit equationy2=−M4(x2−1)−2M2((x−1)x+1)+2M4(−2M2(x−1)+2x−1)+(x−1)2(M2−1)2and is plotted on the right for various values.

The left end of the sector is x=1−M1+M, and the right end is x=1. On the right end, it becomes a cone with angle 2θ where cos⁡θ=1M.

Remarks

As an immediate consequence of this theorem, if z is any nonzero complex number for which the series ∑k=0∞akzk converges, then it follows that limt→1−G(tz)=∑k=0∞akzk in which the limit is taken from below.

The theorem can also be generalized to account for sums which diverge to infinity; see below.Template:Citation needed If ∑k=0∞ak=∞ then limx→1−G(x)→∞.

However, if the series is only known to be divergent, but for reasons other than diverging to infinity, then the claim of the theorem may fail: take, for example, the power series for 11+z.

At z=1 the series is equal to 1−1+1−1+⋯, but 11+1=12.

We also remark the theorem holds for radii of convergence other than R=1: let G(x)=∑k=0∞akxk be a power series with radius of convergence R>0, and suppose the series converges at x=R. Then G(x) is continuous from the left at x=R, that is, limx→R−G(x)=G(R).

Applications

Abel's theorem allows us to evaluate many series in closed form if we can first prove that they converge. For example, consider the sequence ak=(−1)kk+1. The sum ∑k=0∞ak converges by the alternating series test, so we can apply Abel's theorem to the corresponding generating function Ga(z). We obtain Ga(z)=ln⁡(1+z)z,0<z<1, by integrating the uniformly convergent geometric power series term by term on [−z,0]; thus the series ∑k=0∞(−1)kk+1 converges to limz→1−Ga(z)=ln⁡2 by Abel's theorem. Similarly, ∑k=0∞(−1)k2k+1 converges to arctan⁡1=π4.

Ga(z) is called the generating function of the sequence a. Abel's theorem is frequently useful in dealing with generating functions of real-valued and non-negative sequences, such as probability-generating functions. In particular, it is useful in the theory of Galton–Watson processes.

Outline of proof

Source:[2]

Let S=∑k=0∞ak and sn=(∑k=0nak)−S. Then substituting ak=sk−sk−1 and performing a simple manipulation of the series (summation by parts) results in Ga(z)=S+(1−z)∑k=0∞skzk.

Given ε>0, pick n large enough so that |sk|<ε for all k≥n. We have |(1−z)∑k=n∞skzk|≤ε|1−z|∑k=n∞|z|k=ε|1−z||z|n1−|z|<εM when z lies within the given Stolz angle. Whenever z is sufficiently close to 1 we have |(1−z)∑k=0n−1skzk|<ε, so that |Ga(z)−S|<(M+1)ε when z is both sufficiently close to 1 and within the Stolz angle.

Divergent case

To prove the case for ∑k=0∞ak=+∞, let x∈(0,1) and sn=∑k=0nak. Since (sk)k diverges to +∞, we can find, for any M∈ℝ+, a N∈ℕ such that sk>3M for all k≥N. We write, for n>N: ∑k=0nakxk=[(1−x)∑k=0N−1skxk]+{(1−x)[∑k=Nn−1skxk]+snxn} Since the first term on the right hand side vanishes as x→1, we can find an x1∈(0,1) such that it exceeds −M whenever x1<x<1. The second term may be estimated by: (1−x)[∑k=Nn−1skxk]+snxn>3M{(1−x)[∑k=Nn−1xk]+xn}=3MxN. Hence, if we let x2=(2/3)1/N, then for x2<x<1 this exceeds 2M. Combining, we get, for any n>N and max⁡(x1,x2)<x<1: ∑k=0nakxk>M. This establishes: limx→1−lim infn→∞∑k=0nakxk=+∞

In the absence of additional assumptions, the series ∑k=0∞akxk might not converge when |x|<1, hence the use of the limit inferior.

Converses to a theorem like Abel's are called Tauberian theorems: There is no exact converse, but results conditional on some hypothesis. The field of divergent series, and their summation methods, contains many theorems of abelian type and of tauberian type.

Generalizations

In the real case, the functions xk may be replaced by bounded non-negative functions uk(x) monotonically decreasing with k (i.e., uk(x)≥uk+1(x)) such that limx→1−uk(x)=1 for all k.[3] This follows from Abel's uniform convergence test and applying the uniform limit theorem.

See also

Further reading

References

Template:Reflist

  1. ↑ Template:Cite journal
  2. ↑ Ahlfors 1980, pp. 41-42
  3. ↑ Whittaker and Watson, 2021, p. 48