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.

Theorem

Let the Taylor series

<math display=block>G (x) = \sum_{k=0}^\infty a_k x^k</math>

be a power series with real coefficients <math>a_k</math>. Suppose that the series

<math display=block>\sum_{k=0}^\infty a_k</math>

converges.

Then <math>G(x)</math> is continuous from the left at <math>x = 1,</math> that is,

<math display=block>\lim_{x\to 1^-} G(x) = \sum_{k=0}^\infty a_k.</math>

The same theorem holds for complex power series

<math display=block>G(z) = \sum_{k=0}^\infty a_k z^k,</math>

provided that <math>z \to 1</math> entirely within a single Stolz sector, that is, a region of the open unit disk where

<math display=block>|1-z| \leq M(1-|z|)</math>

for some fixed finite <math>M > 1</math>. Without this restriction, the limit may fail to exist: for example, the power series

<math display=block>\sum_{n>0} \frac{z^{3^n}-z^{2\cdot 3^n n</math>

converges to <math>0</math> at <math>z = 1,</math> but is unbounded near any other point of the form <math>e^{\pi i/3^n},</math> so the value at <math>z = 1</math> is not the limit as <math>z</math> tends to 1 in the whole open disk.

Note that the convergence of <math>\sum_{k=0}^\infty a_k</math> implies that the radius of convergence of the power series <math>\sum_{k=0}^{\infty} a_k z^k </math> is at least 1, ensuring convergence for <math>|z|<1</math>.

Also note that by the uniform limit theorem, <math>G(z)</math> is continuous on the real closed interval <math>[0, t]</math> for <math>t < 1,</math> 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 <math>G(z)</math> to <math>[0, 1]</math> is continuous.

Stolz sector

thumb|20 Stolz sectors, for <math>M</math> ranging from 1.01 to 10. The red lines are the tangents to the cone at the right end.

The Stolz sector <math>|1-z|\leq M(1-|z|)</math> has explicit equation<math display="block">y^2 = -\frac{M^4 (x^2 - 1) - 2 M^2 ((x - 1) x + 1) + 2 \sqrt{M^4 (-2 M^2 (x - 1) + 2 x - 1)} + (x - 1)^2}{(M^2 - 1)^2}</math>and is plotted on the right for various values.

The left end of the sector is <math>x = \frac{1-M}{1+M}</math>, and the right end is <math>x=1</math>. On the right end, it becomes a cone with angle <math>2\theta</math> where <math>\cos\theta = \frac{1}{M}</math>.

Remarks

As an immediate consequence of this theorem, if <math>z</math> is any nonzero complex number for which the series

<math display=block>\sum_{k=0}^\infty a_k z^k</math>

converges, then it follows that

<math display=block>\lim_{t\to 1^{- G(tz) = \sum_{k=0}^\infty a_kz^k</math>

in which the limit is taken from below.

The theorem can also be generalized to account for sums which diverge to infinity; see below. If

<math display=block>\sum_{k=0}^\infty a_k = \infty</math>

then

<math display=block>\lim_{x \to 1^{- G(x) \to \infty.</math>

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

<math display=block>\frac{1}{1+z}.</math>

At <math>z = 1</math> the series is equal to <math>1 - 1 + 1 - 1 + \cdots,</math> but <math>\tfrac{1}{1+1} = \tfrac{1}{2}.</math>

We also remark the theorem holds for radii of convergence other than <math>R = 1</math>: let

<math display=block>G(x) = \sum_{k=0}^\infty a_kx^k</math>

be a power series with radius of convergence <math>R > 0,</math> and suppose the series converges at <math>x = R.</math> Then <math>G(x)</math> is continuous from the left at <math>x = R,</math> that is,

<math display=block>\lim_{x\to R^-}G(x) = G(R).</math>

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

<math display=block>a_k = \frac{(-1)^k}{k+1}.</math>

The sum <math>\sum_{k=0}^\infty a_k</math> converges by the alternating series test, so we can apply Abel's theorem to the corresponding generating function <math>G_a(z)</math>. We obtain

<math display=block>G_a(z) = \frac{\ln(1+z)}{z}, \qquad 0 < z < 1,</math>

by integrating the uniformly convergent geometric power series term by term on <math>[-z, 0]</math>; thus the series

<math display=block>\sum_{k=0}^\infty \frac{(-1)^k}{k+1}</math>

converges to <math>\lim \limits_{z \to 1^-} G_a(z) = \ln 2</math> by Abel's theorem. Similarly,

<math display=block>\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}</math>

converges to <math>\arctan 1 = \tfrac{\pi}{4}.</math>

<math>G_a(z)</math> is called the generating function of the sequence <math>a.</math> 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&ndash;Watson processes.

Outline of proof

Source:

Let <math>S = \sum_{k=0}^\infty a_k</math> and <math>s_n= \left({\sum_{k=0}^n a_k}\right) - S\!.</math> Then substituting <math>a_k=s_k-s_{k-1}</math> and performing a simple manipulation of the series (summation by parts) results in

<math display=block>G_a(z) = S + (1-z)\sum_{k=0}^{\infty} s_k z^k.</math>

Given <math>\varepsilon > 0,</math> pick <math>n</math> large enough so that <math>|s_k| < \varepsilon</math> for all <math>k \geq n</math> and note that

<math display=block>\left|(1-z)\sum_{k=n}^\infty s_kz^k \right| \leq \varepsilon |1-z|\sum_{k=n}^\infty |z|^k = \varepsilon|1-z|\frac{|z|^n}{1-|z|} < \varepsilon M </math>

when <math>z</math> lies within the given Stolz angle. Whenever <math>z</math> is sufficiently close to <math>1</math> we have

<math display=block>\left|(1-z)\sum_{k=0}^{n-1} s_kz^k \right| < \varepsilon,</math>

so that <math>\left|G_a(z) - S\right| < (M+1) \varepsilon</math> when <math>z</math> is both sufficiently close to <math>1</math> and within the Stolz angle.

Divergent case

To prove the case for <math>\sum_{k=0}^\infty a_k = +\infty</math>, let <math>x \in (0,1)

</math> and <math> s_n = \sum_{k=0}^n a_k </math>. Since <math>(s_k)_k </math> diverges to <math>+\infty</math>, we can find, for any <math>M \in \mathbb{R}_+ </math>, a <math>N \in \N </math> such that <math>s_k > 3M </math> for all <math>k \geq N

</math>. We write, for <math>n > N</math>:

<math display=block>\sum_{k=0}^n a_k x^k = \left[{(1-x) \sum_{k=0}^{N-1} s_k x^k}\right] + \left\ u_k(x) = 1</math> for all <math>k</math>. This follows from Abel's uniform convergence test and applying the uniform limit theorem.

See also

Further reading

  • - Ahlfors called it Abel's limit theorem.

References

  • (a more general look at Abelian theorems of this type)