Cauchy condensation test
converges if and only if the sum
converges. Moreover, in that case we have
A geometric view is that we are approximating the sum with trapezoids at every . Another explanation is that, as with the analogy between finite sums and integrals, the 'condensation' of terms is analogous to a substitution of an exponential function. This becomes clearer in examples such as
Here the series definitely converges for a > 1, and diverges for a < 1. When a = 1, the condensation transformation essentially gives the series
The logarithms 'shift to the left'. So when a = 1, we have convergence for b > 1, divergence for b < 1. When b = 1 the value of c enters.
Let f(n) be a positive, non-increasing sequence of real numbers. To simplify the notation, we will write an = f(n). We are to investigate the series . The condensation test follows from noting that if we collect the terms of the series into groups of lengths , each of these groups will be less than by monotonicity. Observe,
We have used the fact that the sequence an is non-increasing, thus whenever . The convergence of the original series now follows from direct comparison to this "condensed" series. To see that convergence of the original series implies the convergence of this last series, we similarly put,
And we have convergence, again by direct comparison. And we are done. Note that we have obtained the estimate
The following generalization is due to Schlömilch. Let be an infinite real series whose terms are positive and non-increasing, and let be a strictly increasing sequence of positive integers such that
is bounded, where is the forward difference. Then the series converges if the series
Taking , we see , so the Cauchy condensation test emerges as a special case.
- Bonar, Khoury (2006). Real Infinite Series. Mathematical Association of America. ISBN 0-88385-745-6.