# Abel's test

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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, ${\displaystyle \{a_{n}\}}$ and ${\displaystyle \{b_{n}\}}$, if the sequences satisfy

• ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$ converges
• ${\displaystyle \lbrace b_{n}\rbrace \,}$ is monotonic and ${\displaystyle \lim _{n\rightarrow \infty }b_{n}\neq \infty }$

then the series

${\displaystyle \sum _{n=1}^{\infty }a_{n}b_{n}}$

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

${\displaystyle \lim _{n\rightarrow \infty }a_{n}=0\,}$

and the series

${\displaystyle f(z)=\sum _{n=0}^{\infty }a_{n}z^{n}\,}$

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.[1]

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

${\displaystyle z=e^{i\theta }\quad \Rightarrow \quad z^{\frac {1}{2}}-z^{-{\frac {1}{2}}}=2i\sin {\textstyle {\frac {\theta }{2}}}\neq 0}$

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

{\displaystyle {\begin{aligned}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_{p}z^{p+{\frac {1}{2}}}\,\end{aligned}}}

where Sp and Sq are partial sums:

${\displaystyle S_{p}=\sum _{n=0}^{p}a_{n}z^{n}.\,}$

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

{\displaystyle {\begin{aligned}\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|\\&\leq \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_{p}z^{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{aligned}}}

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.