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# Abel's test

In mathematics, Abel's test (also known as Abel's criterion) is a method of testing for the convergence of an infinite series. The test is named after mathematician Niels Abel. There are two slightly different versions of Abel's test – one is used with series of real numbers, and the other is used with power series in complex analysis.

Abel's test in real analysis

If the following statements are true:

1. \( \sum a_n \) is a convergent series

2. \( {b_n} \) is a monotone sequence

3. \( {b_n} \) is bounded,

then \sum a_nb_n is also convergent.

Abel's test in complex analysis

A closely related convergence test, also known as Abel's test, can often be used to establish the convergence of a power series on the boundary of its circle of convergence. Specifically, Abel's test states that if

\( \lim_{n\rightarrow\infty} a_n = 0\, \)

and the series

\( f(z) = \sum_{n=0}^\infty a_nz^n\, \)

converges when |z| < 1 and diverges when |z| > 1, and the coefficients \( {a_n} \) are positive real numbers decreasing monotonically toward the limit zero for n > m (for large enough n, in other words), then the power series for f(z) converges everywhere on the unit circle, except when z = 1. Abel's test cannot be applied when z = 1, so convergence at that single point must be investigated separately. Notice that Abel's test can also be applied to a power series with radius of convergence R ≠ 1 by a simple change of variables ζ = z/R.[1]

Proof of Abel's test: Suppose that z is a point on the unit circle, z ≠ 1. Then

\( z = e^{i\theta} \quad\Rightarrow\quad z^{\frac{1}{2}} - z^{-\frac{1}{2}} = 2i\sin{\textstyle \frac{\theta}{2}} \ne 0 \)

so that, for any two positive integers p > q > m, we can write

\( \begin{align} 2i\sin{\textstyle \frac{\theta}{2}}\left(S_p - S_q\right) & = \sum_{n=q+1}^p a_n \left(z^{n+\frac{1}{2}} - z^{n-\frac{1}{2}}\right)\\ & = \left[\sum_{n=q+2}^p \left(a_{n-1} - a_n\right) z^{n-\frac{1}{2}}\right] - a_{q+1}z^{q+\frac{1}{2}} + a_pz^{p+\frac{1}{2}}\, \end{align} \)

where Sp and Sq are partial sums:

\( S_p = \sum_{n=0}^p a_nz^n.\, \)

But now, since |z| = 1 and the an are monotonically decreasing positive real numbers when n > m, we can also write

\( \begin{align} \left| 2i\sin{\textstyle \frac{\theta}{2}}\left(S_p - S_q\right)\right| & = \left| \sum_{n=q+1}^p a_n \left(z^{n+\frac{1}{2}} - z^{n-\frac{1}{2}}\right)\right| \\ & \le \left[\sum_{n=q+2}^p \left| \left(a_{n-1} - a_n\right) z^{n-\frac{1}{2}}\right|\right] + \left| a_{q+1}z^{q+\frac{1}{2}}\right| + \left| a_pz^{p+\frac{1}{2}}\right| \\ & = \left[\sum_{n=q+2}^p \left(a_{n-1} - a_n\right)\right] +a_{q+1} + a_p \\ & = a_{q+1} - a_p + a_{q+1} + a_p = 2a_{q+1}.\, \end{align} \)

Now we can apply Cauchy's criterion to conclude that the power series for f(z) converges at the chosen point z ≠ 1, because sin(½θ) ≠ 0 is a fixed quantity, and aq+1 can be made smaller than any given ε > 0 by choosing a large enough q.

Notes

^ (Moretti, 1964, p. 91)

References

Gino Moretti, Functions of a Complex Variable, Prentice-Hall, Inc., 1964

See also

Abel's uniform convergence test

External links

Proof (for real series) at PlanetMath.org

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