Integral test for convergence

In mathematics, the integral test for convergence is a method used to test infinite series of monotonic terms for convergence.

It was developed by Colin Maclaurin and Augustin-Louis Cauchy and is sometimes known as the Maclaurin–Cauchy test.

Consider an integer N and a function f defined on the unbounded interval [N, ∞), on which it is monotone decreasing.

Then the infinite series converges to a real number if and only if the improper integral is finite.

If the improper integral is finite, then the proof also gives the lower and upper bounds for the infinite series.

is decreasing and the above theorem applies.

[4][better source needed] The proof uses the comparison test, comparing the term

Note that this set contains an open non-empty interval precisely if

as the rational number that has the least index in an enumeration

is monotone, this defines an injective mapping

This is sufficient for Riemann integrability.

[5] Since f is a monotone decreasing function, we know that and Hence, for every integer n ≥ N, and, for every integer n ≥ N + 1, By summation over all n from N to some larger integer M, we get from (2) and from (3) Combining these two estimates yields Letting M tend to infinity, the bounds in (1) and the result follow.

The harmonic series diverges because, using the natural logarithm, its antiderivative, and the fundamental theorem of calculus, we get On the other hand, the series (cf.

Riemann zeta function) converges for every ε > 0, because by the power rule From (1) we get the upper estimate which can be compared with some of the particular values of Riemann zeta function.

The above examples involving the harmonic series raise the question of whether there are monotone sequences such that f(n) decreases to 0 faster than 1/n but slower than 1/n1+ε in the sense that for every ε > 0, and whether the corresponding series of the f(n) still diverges.

Once such a sequence is found, a similar question can be asked with f(n) taking the role of 1/n, and so on.

In this way it is possible to investigate the borderline between divergence and convergence of infinite series.

Using the integral test for convergence, one can show (see below) that, for every natural number k, the series still diverges (cf.

proof that the sum of the reciprocals of the primes diverges for k = 1) but converges for every ε > 0.

Here lnk denotes the k-fold composition of the natural logarithm defined recursively by Furthermore, Nk denotes the smallest natural number such that the k-fold composition is well-defined and lnk(Nk) ≥ 1, i.e. using tetration or Knuth's up-arrow notation.

To see the divergence of the series (4) using the integral test, note that by repeated application of the chain rule hence To see the convergence of the series (5), note that by the power rule, the chain rule and the above result hence and (1) gives bounds for the infinite series in (5).