Abel's test

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In mathematics, Abel's test 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.

1 Abel's test in real analysis
2 Abel's test in complex analysis
3 External links
4 Notes
5 References

[edit] Abel's test in real analysis

Given two sequences of real numbers, ${a n}$ and ${b n}$ , if the sequences satisfy

$\sum^{\infty}_{n=1}a_n {\ne} {\infty}$

$\lbrace b_n \rbrace\,$ is monotonic and $\lim_{n \rightarrow \infty} b_n \ne \infty$

then the series

$\sum^{\infty}_{n=1}a_n b_n$

converges.

[edit] 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]

Here's why Abel's test works. 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 S_p and S_q are partial sums:

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

But now, since |z| = 1 and the a_n 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 a_q+1 can be made smaller than any given ε > 0 by choosing a large enough q.