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.
Statement of the test Consider an integer N and a function f defined on the unbounded interval [N, ∞), on which it is monotonically decreasing. Then the infinite series
∑ n = N ∞ f ( n ) {\displaystyle \sum _{n=N}^{\infty }f(n)}
converges to a real number if and only if the improper integral
∫ N ∞ f ( x ) d x {\displaystyle \int _{N}^{\infty }f(x)\,dx}
is finite. In particular, if the integral diverges, then the series diverges as well.
Remark If the improper integral is finite, then the proof also gives the lower and upper bounds
for the infinite series. Note that if the function f ( x ) {\displaystyle f(x)} is increasing, then the function − f ( x ) {\displaystyle -f(x)} is decreasing and the above theorem applies. Many textbooks require the function f {\displaystyle f} to be positive, but this condition is not really necessary, since when f {\displaystyle f} is negative and decreasing both ∑ n = N ∞ f ( n ) {\displaystyle \sum _{n=N}^{\infty }f(n)} and ∫ N ∞ f ( x ) d x {\displaystyle \int _{N}^{\infty }f(x)\,dx} diverge.
Proof The proof uses the comparison test, comparing the term f ( n ) {\displaystyle f(n)} with the integral of f {\displaystyle f} over the intervals [ n − 1 , n ) {\displaystyle [n-1,n)} and [ n , n + 1 ) {\displaystyle [n,n+1)} respectively. The monotonic function f {\displaystyle f} is continuous almost everywhere. To show this, let
D = { x ∈ [ N , ∞ ) ∣ f is discontinuous at x } {\displaystyle D=\{x\in [N,\infty )\mid f{\text{ is discontinuous at }}x\}}
For every x ∈ D {\displaystyle x\in D} , there exists by the density of Q {\displaystyle \mathbb {Q} } , a c ( x ) ∈ Q {\displaystyle c(x)\in \mathbb {Q} } so that c ( x ) ∈ [ lim y ↓ x f ( y ) , lim y ↑ x f ( y ) ] {\displaystyle c(x)\in \left[\lim _{y\downarrow x}f(y),\lim _{y\uparrow x}f(y)\right]} . Note that this set contains an open non-empty interval precisely if f {\displaystyle f} is discontinuous at x {\displaystyle x} . We can uniquely identify c ( x ) {\displaystyle c(x)} as the rational number that has the least index in an enumeration N → Q {\displaystyle \mathbb {N} \to \mathbb {Q} } and satisfies the above property. Since f {\displaystyle f} is monotone, this defines an injective mapping c : D → Q , x ↦ c ( x ) {\displaystyle c:D\to \mathbb {Q} ,x\mapsto c(x)} and thus D {\displaystyle D} is countable. It follows that f {\displaystyle f} is continuous almost everywhere. This is sufficient for Riemann integrability. Since f is a monotone decreasing function, we know that
f ( x ) ≤ f ( n ) for all x ∈ [ n , ∞ ) {\displaystyle f(x)\leq f(n)\quad {\text{for all }}x\in [n,\infty )}
and
f ( n ) ≤ f ( x ) for all x ∈ [ N , n ] . {\displaystyle f(n)\leq f(x)\quad {\text{for all }}x\in [N,n].}
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)
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