Integral test for convergence

The integral test applied to the harmonic series. Since the area under the curve Template:Math for Template:Math is infinite, the total area of the rectangles must be infinite as well.

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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 Template:Math and a function Template:Math defined on the unbounded interval Template:Closed-open, on which it is monotone decreasing. Then the infinite series

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

converges to a real number if and only if the improper integral

$\underset{N}{\overset{\mathrm{\infty }}{\int }}f(x)dx$

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 $f(x)$ is increasing, then the function $-f(x)$ is decreasing and the above theorem applies.

Many textbooks require the function $f$ to be positive, but this condition is not really necessary, since when $f$ is negative and decreasing both $\sum _{n=N}^{\mathrm{\infty }}f(n)$ and $\underset{N}{\overset{\mathrm{\infty }}{\int }}f(x)dx$ diverge.Template:Better source needed

Proof

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

The monotonic function $f$ is continuous almost everywhere. To show this, let

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

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

Note that this set contains an open non-empty interval precisely if $f$ is discontinuous at $x$. We can uniquely identify $c(x)$ as the rational number that has the least index in an enumeration $\mathbb{N}\to \mathbb{Q}$ and satisfies the above property. Since $f$ is monotone, this defines an injective mapping $c:D\to \mathbb{Q},x\mapsto c(x)$ and thus $D$ is countable. It follows that $f$ is continuous almost everywhere. This is sufficient for Riemann integrability.

Since Template:Math is a monotone decreasing function, we know that

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

and

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

Hence, for every integer Template:Math,

and, for every integer Template:Math,

By summation over all Template:Math from Template:Math to some larger integer Template:Math, we get from (Template:EquationNote)

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

and from (Template:EquationNote)

$\begin{array}{cc}\sum _{n=N}^{M}f(n)& =f(N)+\sum _{n=N+1}^{M}f(n)\\ & \le f(N)+\sum _{n=N+1}^M\underset{\ge f(n)}{\underbrace{\underset{n-1}{\overset{n}{\int }}f(x)dx}}\\ & =f(N)+\underset{N}{\overset{M}{\int }}f(x)dx.\end{array}$

Combining these two estimates yields

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

Letting Template:Math tend to infinity, the bounds in (Template:EquationNote) and the result follow.

Applications

The harmonic series

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

diverges because, using the natural logarithm, its antiderivative, and the fundamental theorem of calculus, we get

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

On the other hand, the series

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

(cf. Riemann zeta function) converges for every Template:Math, because by the power rule

$\underset{1}{\overset{M}{\int }}\frac{1}{n^{1+\epsilon }}dn={-\frac{1}{\epsilon n^{\epsilon }}|}_1^M=\frac{1}{\epsilon }(1-\frac{1}{M^{\epsilon }})\le \frac{1}{\epsilon }<\mathrm{\infty }\text{for all }M\ge 1.$

From (Template:EquationNote) we get the upper estimate

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

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

Borderline between divergence and convergence

The above examples involving the harmonic series raise the question of whether there are monotone sequences such that Template:Math decreases to 0 faster than Template:Math but slower than Template:Math in the sense that

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

for every Template:Math, and whether the corresponding series of the Template:Math still diverges. Once such a sequence is found, a similar question can be asked with Template:Math taking the role of Template:Math, 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 Template:Math, the series Template:NumBlk still diverges (cf. proof that the sum of the reciprocals of the primes diverges for Template:Math) but Template:NumBlk converges for every Template:Math. Here Template:Math denotes the Template:Math-fold composition of the natural logarithm defined recursively by

${\ln }_k(x)=\{\begin{array}{cc}\ln (x)& \text{for }k=1,\\ \ln ({\ln }_{k-1}(x))& \text{for }k\ge 2.\end{array}$

Furthermore, Template:Math denotes the smallest natural number such that the Template:Math-fold composition is well-defined and Template:Math, i.e.

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

using tetration or Knuth's up-arrow notation.

To see the divergence of the series (Template:EquationNote) using the integral test, note that by repeated application of the chain rule

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

hence

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

To see the convergence of the series (Template:EquationNote), note that by the power rule, the chain rule and the above result

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

hence

$\underset{N_k}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x\ln (x)\cdots {\ln }_{k-1}(x)({\ln }_k(x){)}^{1+\epsilon }}=-\frac{1}{\epsilon ({\ln }_k(x){)}^{\epsilon }}{|}_{N_k}^{\mathrm{\infty }}<\mathrm{\infty }$

and (Template:EquationNote) gives bounds for the infinite series in (Template:EquationNote).

See also

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

References

• Knopp, Konrad, "Infinite Sequences and Series", Dover Publications, Inc., New York, 1956. (§ 3.3) Template:ISBN
• Whittaker, E. T., and Watson, G. N., A Course in Modern Analysis, fourth edition, Cambridge University Press, 1963. (§ 4.43) Template:ISBN
• Ferreira, Jaime Campos, Ed Calouste Gulbenkian, 1987, Template:ISBN