Raabe’s convergence test

The ratio test for the convergence of a series is inconclusive if the ratio of consecutive terms converges to 1. There are more advanced variations on the ratio test, such as Raabe’s test than may succeed when the basic ratio test fails.

For example, consider the hypergeometric function F(11, 3; 21; z) with parameters taken from today’s date: 11/3/21. Since hypergeometric functions are symmetric in their first two parameters, European readers are welcome to think of this function as F(3, 11; 21; z).

F(11, 3; 21; z) = 1 + \frac{11\cdot 3}{21\cdot 1}z + \frac{11\cdot 12 \cdot 3\cdot 4}{21\cdot 22\cdot 1\cdot 2}z^2 + \cdots

The coefficient of zk is

(11)k (3)k / (21)k (1)k


(a)k = a (a+1) (a + 2) … (a + k – 1)

is the kth rising power of a.

The series defining F(11, 3; 21; z) converges uniformly if |z| < 1 and diverges if |z| > 1. What happens on the unit circle where |z| = 1?

The ratio of the absolute value of consecutive terms is

\frac{a_n}{a_{n+1}} = \frac{(n+21)(n+1)}{(n+11)(n+3)} = 1 + \frac{8}{n} - {\cal O}\left(\frac{1}{n^2} \right)

which converges to 1 as n → ∞. The basic ratio test is inconclusive.

Raabe’s ratio test says to consider

\rho_n = n\left(\frac{a_n}{a_{n+1}} - 1 \right)

If the limit (more generally the lim inf) of ρn is greater than 1, the series converges. if the limit (more generally the lim sup) of ρn is less than 1, the series diverges.

In our case the limit of ρn is 8, and so F(11, 3; 21; z) converges everywhere on the unit circle.

Like the basic ratio test, Raabe’s test may also be inconclusive. In that case there are more powerful tests to try, such as Kummer’s test, though it and all related tests may turn out inconclusive.

By the way Mr. Raabe’s name came up here a couple weeks ago in the context of Bernoulli polynomials.