To see the first inequality, the terms of the original series are rebracketed into runs whose lengths are powers of two, and then each run is bounded above by replacing each term by the largest term in that run: the first one, since the terms are non-increasing.
To see the second, the two series are again rebracketed into runs of power of two length, but "offset" as shown below, so that the run of which begins with lines up with the end of the run of which ends with , so that the former stays always "ahead" of the latter.
Visualization of the above argument. Partial sums of the series , , and are pictured.
The "condensation" transformation recalls the integral variable substitution yielding .
Pursuing this idea, the integral test for convergence gives us that converges if and only if converges. The substitution yields the integral and another integral test brings us to the condensed series .
The test can be useful for series where n appears as in a denominator in f. For the most basic example of this sort, the harmonic series is transformed into the series , which clearly diverges.
As a more complex example, take
Here the series definitely converges for a > 1, and diverges for a < 1. When a = 1, the condensation transformation 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.
The condensation test, applied repeatedly, can be used to show that for any , the generalized Bertrand series
converges for and diverges for . Here, we have used the notation for the mth iterate of a function , so that with . The lower limit of the sum was chosen so that all terms of the series are positive. Notably, this result provides examples of series that converge or diverge at an arbitrarily slow rate. For instance, when and , the sum exceeds 10 only after a googolplex terms yet diverges nevertheless.