Abel's Test - Abel's Test in Complex Analysis

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 -
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 +
\left| a_{q+1}z^{q+\frac{1}{2}}\right| + \left| a_pz^{p+\frac{1}{2}}\right| \\
& = \left +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.

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