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
Given two sequences of real numbers, $$\{a_n\}$$ and $$\{b_n\}$$, if the sequences satisfy


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


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

then the series


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

converges.

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 {an} 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.

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.