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Inmathematics, theintegral test for convergence is amethod used to test infiniteseries ofmonotonic terms forconvergence. It was developed byColin Maclaurin andAugustin-Louis Cauchy and is sometimes known as theMaclaurin–Cauchy test.

Consider anintegerN and a functionf defined on the unboundedinterval[N, ∞), on which it ismonotone decreasing. Then the infinite series

${\displaystyle \sum _{n=N}^{\mathrm{\infty }}f(n)}$

converges to areal number if and only if theimproper integral

${\displaystyle {\int }_N^{\mathrm{\infty }}f(x)dx}$

is finite. In particular, if the integral diverges, then theseries diverges as well.

If the improper integral is finite, then the proof also gives thelower and upper bounds

${\displaystyle {\int }_N^{\mathrm{\infty }}f(x)dx\le \sum _{n=N}^{\mathrm{\infty }}f(n)\le f(N)+{\int }_N^{\mathrm{\infty }}f(x)dx}$

1

for the infinite series.

Note that if the function${\displaystyle f(x)}$ is increasing, then the function${\displaystyle -f(x)}$ is decreasing and the above theorem applies.

Many textbooks require the function${\displaystyle f}$ to be positive,^[1][2][3] but this condition is not really necessary, since when${\displaystyle f}$ is negative and decreasing both${\displaystyle \sum _{n=N}^{\mathrm{\infty }}f(n)}$ and${\displaystyle {\int }_N^{\mathrm{\infty }}f(x)dx}$ diverge.^{[4][better source needed]}

The proof uses thecomparison test, comparing the term${\displaystyle f(n)}$ with the integral of${\displaystyle f}$ over the intervals${\displaystyle [n-1,n)}$ and${\displaystyle [n,n+1)}$ respectively.

The monotonic function${\displaystyle f}$ iscontinuousalmost everywhere. To show this, let

${\displaystyle D=\{x\in [N,\mathrm{\infty })\mid f\text{ is discontinuous at }x\}}$

For every${\displaystyle x\in D}$, there exists by thedensity of${\displaystyle \mathbb{Q}}$, a${\displaystyle c(x)\in \mathbb{Q}}$ so that${\displaystyle c(x)\in \left[\underset{y\downarrow x}{lim}f(y),\underset{y\uparrow x}{lim}f(y)\right]}$.

Note that this set contains anopennon-empty interval precisely if${\displaystyle f}$ isdiscontinuous at${\displaystyle x}$. We can uniquely identify${\displaystyle c(x)}$ as therational number that has the least index in anenumeration${\displaystyle \mathbb{N}\to \mathbb{Q}}$ and satisfies the above property. Since${\displaystyle f}$ ismonotone, this defines aninjectivemapping${\displaystyle c:D\to \mathbb{Q},x\mapsto c(x)}$ and thus${\displaystyle D}$ iscountable. It follows that${\displaystyle f}$ iscontinuousalmost everywhere. This issufficient forRiemann integrability.^[5]

Sincef is a monotone decreasing function, we know that

${\displaystyle f(x)\le f(n)\text{for all }x\in [n,\mathrm{\infty })}$

and

${\displaystyle f(n)\le f(x)\text{for all }x\in [N,n].}$

Hence, for every integern ≥N,

${\displaystyle {\int }_n^{n+1}f(x)dx\le {\int }_n^{n+1}f(n)dx=f(n)}$

2

and, for every integern ≥N + 1,

${\displaystyle f(n)={\int }_{n-1}^{n}f(n)dx\le {\int }_{n-1}^{n}f(x)dx.}$

3

By summation over alln fromN to some larger integerM, we get from (2)

${\displaystyle {\int }_N^{M+1}f(x)dx=\sum _{n=N}^M\underset{\le f(n)}{\underbrace{{\int }_n^{n+1}f(x)dx}}\le \sum _{n=N}^{M}f(n)}$

and from (3)

Combining these two estimates yields

${\displaystyle {\int }_N^{M+1}f(x)dx\le \sum _{n=N}^{M}f(n)\le f(N)+{\int }_N^{M}f(x)dx.}$

LettingM tend to infinity, the bounds in (1) and the result follow.

Theharmonic series

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}}$

diverges because, using thenatural logarithm, itsantiderivative, and thefundamental theorem of calculus, we get

${\displaystyle {\int }_1^M\frac{1}{n}dn=\ln n{|}_1^M=\ln M\to \mathrm{\infty }\text{for }M\to \mathrm{\infty }.}$

On the other hand, the series

${\displaystyle \zeta (1+\epsilon )=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^{1+\epsilon }}}$

(cf.Riemann zeta function)converges for everyε > 0, because by thepower rule

From (1) we get the upper estimate

${\displaystyle \zeta (1+\epsilon )=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^{1+\epsilon }}\le \frac{1+\epsilon }{\epsilon },}$

which can be compared with some of theparticular values of Riemann zeta function.

The above examples involving the harmonic series raise the question of whether there are monotone sequences such thatf(n) decreases to 0 faster than1/n but slower than1/n^1+ε in the sense that

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{f(n)}{1/n}=0\text{and}\underset{n\to \mathrm{\infty }}{lim}\frac{f(n)}{1/n^{1+\epsilon }}=\mathrm{\infty }}$

for everyε > 0, and whether the corresponding series of thef(n) still diverges. Once such a sequence is found, a similar question can be asked withf(n) taking the role of1/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 everynatural numberk, the series

${\displaystyle \sum _{n=N_k}^{\mathrm{\infty }}\frac{1}{n\ln (n){\ln }_2(n)\cdots {\ln }_{k-1}(n){\ln }_k(n)}}$

4

still diverges (cf.proof that the sum of the reciprocals of the primes diverges fork = 1) but

${\displaystyle \sum _{n=N_k}^{\mathrm{\infty }}\frac{1}{n\ln (n){\ln }_2(n)\cdots {\ln }_{k-1}(n)({\ln }_k(n){)}^{1+\epsilon }}}$

5

converges for everyε > 0. Hereln_k denotes thek-foldcomposition of the natural logarithm definedrecursively by

Furthermore,N_k denotes the smallest natural number such that thek-fold composition is well-defined andln_k(N_k) ≥ 1, i.e.

${\displaystyle N_k\ge \underset{k{e}^{\prime }\text{s}}{\underbrace{e^{e^{{\cdot }^{{\cdot }^e}}}}}=e\uparrow \uparrow k}$

usingtetration orKnuth's up-arrow notation.

To see the divergence of the series (4) using the integral test, note that by repeated application of thechain rule

${\displaystyle \frac{d}{dx}{\ln }_{k+1}(x)=\frac{d}{dx}\ln ({\ln }_k(x))=\frac{1}{{\ln }_k(x)}\frac{d}{dx}{\ln }_k(x)=\cdots =\frac{1}{x\ln (x)\cdots {\ln }_k(x)},}$

hence

${\displaystyle {\int }_{N_k}^{\mathrm{\infty }}\frac{dx}{x\ln (x)\cdots {\ln }_k(x)}={\ln }_{k+1}(x){|}_{N_k}^{\mathrm{\infty }}=\mathrm{\infty }.}$

To see the convergence of the series (5), note that by thepower rule, the chain rule and the above result

${\displaystyle -\frac{d}{dx}\frac{1}{\epsilon ({\ln }_k(x){)}^{\epsilon }}=\frac{1}{({\ln }_k(x){)}^{1+\epsilon }}\frac{d}{dx}{\ln }_k(x)=\cdots =\frac{1}{x\ln (x)\cdots {\ln }_{k-1}(x)({\ln }_k(x){)}^{1+\epsilon }},}$

hence

and (1) gives bounds for the infinite series in (5).

• Convergence tests
• Convergence (mathematics)
• Direct comparison test
• Dominated convergence theorem
• Euler-Maclaurin formula
• Limit comparison test
• Monotone convergence theorem

• Knopp, Konrad, "Infinite Sequences and Series",Dover Publications, Inc., New York, 1956. (§ 3.3)ISBN 0-486-60153-6
• Whittaker, E. T., and Watson, G. N.,A Course in Modern Analysis, fourth edition, Cambridge University Press, 1963. (§ 4.43)ISBN 0-521-58807-3
• Ferreira, Jaime Campos, Ed Calouste Gulbenkian, 1987,ISBN 972-31-0179-3

• Augustin-Louis Cauchy
• Integral calculus
• Convergence tests

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