View source for Integral test for convergence

{{Short description|Test for infinite series of monotonous terms for convergence}}
[[File:Integral Test.svg|thumb|right|300px|The integral test applied to the [[Harmonic series (mathematics)|harmonic series]].  Since the area under the curve {{math|''y'' {{=}} 1/''x''}} for {{math|''x'' ∈ {{closed-open|1, ∞}}}} is infinite, the total area of the rectangles must be infinite as well.]]
{{Calculus|Series}}

In [[Mathematics|mathematics]], the '''integral test for convergence''' is a [[Convergence tests|method used to test]] infinite [[Series (mathematics)|series]] of [[Monotonic function|monotonous]] terms for [[Convergent series|convergence]]. It was developed by Colin Maclaurin and [[Biography:Augustin-Louis Cauchy|Augustin-Louis Cauchy]] and is sometimes known as the '''Maclaurin–Cauchy test'''.

==Statement of the test==
Consider an [[Integer|integer]] {{math|''N''}} and a function {{math|''f''}} defined on the unbounded [[Interval (mathematics)|interval]] {{closed-open|''N'', ∞}}, on which it is monotone decreasing. Then the infinite series

:<math>\sum_{n=N}^\infty f(n)</math>

converges to a [[Real number|real number]] if and only if the [[Improper integral|improper integral]]

:<math>\int_N^\infty f(x)\,dx</math>

is finite. In particular, if the integral diverges, then the [[Divergent series|series diverges]] as well.

===Remark===
If the improper integral is finite, then the proof also gives the [[Upper and lower bounds|lower and upper bounds]]

{{NumBlk|:|<math>\int_N^\infty f(x)\,dx\le\sum_{n=N}^\infty f(n)\le f(N)+\int_N^\infty f(x)\,dx</math>|{{EquationRef|1}}}}

for the infinite series.

Note that if the function <math>f(x)</math> is increasing, then the function <math>-f(x)</math> is decreasing and the above theorem applies.

==Proof==
The proof basically uses the [[Direct comparison test|comparison test]], comparing the term {{math|''f''(''n'')}} with the integral of {{math|''f''}} over the intervals
{{closed-open|''n'' − 1, ''n''}} and {{closed-open|''n'', ''n'' + 1}}, respectively.

The monotonous function <math>f</math> is [[Continuous function|continuous]] [[Almost everywhere|almost everywhere]]. To show this, let <math>D=\{ x\in [N,\infty)\mid f\text{ is discontinuous at } x\}</math>. For every <math>x\in D</math>, there exists by the [[Dense set|density]] of <math>\mathbb Q</math> a <math>c(x)\in\mathbb Q</math> so that <math>c(x)\in\left[\lim_{y\downarrow x} f(y), \lim_{y\uparrow x} f(y)\right]</math>. Note that this set contains an [[Open set|open]] non-empty interval precisely if <math>f</math> is discontinuous at <math>x</math>. We can uniquely identify <math>c(x)</math> as the [[Rational number|rational number]] that has the least index in an [[Enumeration|enumeration]] <math>\mathbb N\to\mathbb Q</math> and satisfies the above property. Since <math>f</math> is [[Monotonic function|monotone]], this defines an [[Injective function|injective]] [[Function (mathematics)|mapping]] <math>c:D\to\mathbb Q, x\mapsto c(x)</math> and thus <math>D</math> is [[Countable set|countable]]. It follows that <math>f</math> is [[Continuous function|continuous]] [[Almost everywhere|almost everywhere]]. This is [[Necessity and sufficiency|sufficient]] for Riemann integrability.<ref>{{Cite journal| issn = 0002-9890| volume = 43
| issue = 7| pages = 396–398| last = Brown| first = A. B.| title = A Proof of the Lebesgue Condition for Riemann Integrability| journal = The American Mathematical Monthly| date = September 1936| jstor = 2301737| doi = 10.2307/2301737}}</ref>

Since {{math|''f''}} is a monotone decreasing function, we know that

:<math>
f(x)\le f(n)\quad\text{for all }x\in[n,\infty)
</math>

and

:<math>
f(n)\le f(x)\quad\text{for all }x\in[N,n].
</math>

Hence, for every integer {{math|''n'' ≥ ''N''}},

{{NumBlk|:|<math>
\int_n^{n+1} f(x)\,dx
\le\int_{n}^{n+1} f(n)\,dx
=f(n)</math>|{{EquationRef|2}}}}

and, for every integer {{math|''n'' ≥ ''N'' + 1}},

{{NumBlk|:|<math>
f(n)=\int_{n-1}^{n} f(n)\,dx
\le\int_{n-1}^n f(x)\,dx.
</math>|{{EquationRef|3}}}}

By summation over all {{math|''n''}} from {{math|''N''}} to some larger integer {{math|''M''}}, we get from ({{EquationNote|2}})

:<math>
\int_N^{M+1}f(x)\,dx=\sum_{n=N}^M\underbrace{\int_n^{n+1}f(x)\,dx}_{\le\,f(n)}\le\sum_{n=N}^Mf(n)
</math>


:<math>
\sum_{n=N}^Mf(n)=f(N)+\sum_{n=N+1}^Mf(n)\le f(N)+\sum_{n=N+1}^M\underbrace{\int_{n-1}^n f(x)\,dx}_{\ge\,f(n)}=f(N)+\int_N^M f(x)\,dx.
</math>

Combining these two estimates yields

:<math>\int_N^{M+1}f(x)\,dx\le\sum_{n=N}^Mf(n)\le f(N)+\int_N^M f(x)\,dx.</math>

Letting {{math|''M''}} tend to infinity, the bounds in ({{EquationNote|1}}) and the result follow.

==Applications==
The [[Harmonic series (mathematics)|harmonic series]]
:<math>
\sum_{n=1}^\infty \frac 1 n
</math>
diverges because, using the [[Natural logarithm|natural logarithm]], its [[Antiderivative|antiderivative]], and the [[Fundamental theorem of calculus|fundamental theorem of calculus]], we get
:<math>
\int_1^M \frac 1 n\,dn = \ln n\Bigr|_1^M = \ln M \to\infty
\quad\text{for }M\to\infty.
</math>
On the other hand, the series
:<math>
\zeta(1+\varepsilon)=\sum_{n=1}^\infty \frac1{n^{1+\varepsilon}}
</math>
(cf. [[Riemann zeta function]])
converges for every {{math|''ε'' > 0}}, because by the [[Power rule|power rule]]
:<math>
\int_1^M\frac1{n^{1+\varepsilon}}\,dn
= \left. -\frac 1{\varepsilon n^\varepsilon} \right|_1^M=
\frac 1 \varepsilon \left(1-\frac 1 {M^\varepsilon}\right)
\le \frac 1 \varepsilon  < \infty
\quad\text{for all }M\ge1.
</math>
From ({{EquationNote|1}}) we get the upper estimate
:<math>
\zeta(1+\varepsilon)=\sum_{x=1}^\infty \frac 1 {n^{1+\varepsilon}} \le \frac{1 + \varepsilon}\varepsilon,
</math>
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 {{math|''f''(''n'')}} decreases to 0 faster than {{math|1/''n''}} but slower than {{math|1/''n''<sup>1+''ε''</sup>}} in the sense that
:<math>
\lim_{n\to\infty}\frac{f(n)}{1/n}=0
\quad\text{and}\quad
\lim_{n\to\infty}\frac{f(n)}{1/n^{1+\varepsilon}}=\infty
</math>
for every {{math|''ε'' > 0}}, and whether the corresponding series of the {{math|''f''(''n'')}} still diverges. Once such a sequence is found, a similar question can be asked with {{math|''f''(''n'')}} taking the role of {{math|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|natural number]] {{math|''k''}}, the series
{{NumBlk|:|<math>
\sum_{n=N_k}^\infty\frac1{n\ln(n)\ln_2(n)\cdots \ln_{k-1}(n)\ln_k(n)}
</math>|{{EquationRef|4}}}}
still diverges (cf. proof that the sum of the reciprocals of the primes diverges for {{math|''k'' {{=}} 1}}) but
{{NumBlk|:|<math>
\sum_{n=N_k}^\infty\frac1{n\ln(n)\ln_2(n)\cdots\ln_{k-1}(n)(\ln_k(n))^{1+\varepsilon}}
</math>|{{EquationRef|5}}}}
converges for every {{math|''ε'' > 0}}. Here {{math|ln<sub>''k''</sub>}} denotes the {{math|''k''}}-fold [[Function composition|composition]] of the natural logarithm defined [[Recursion|recursively]] by
:<math>
\ln_k(x)=
\begin{cases}
\ln(x)&\text{for }k=1,\\
\ln(\ln_{k-1}(x))&\text{for }k\ge2.
\end{cases}
</math>
Furthermore, {{math|''N''<sub>''k''</sub>}} denotes the smallest natural number such that the {{math|''k''}}-fold composition is well-defined and {{math|ln<sub>''k''</sub>(''N''<sub>''k''</sub>) ≥ 1}}, i.e.
:<math>
N_k\ge \underbrace{e^{e^{\cdot^{\cdot^{e}}}}}_{k\ e'\text{s}}=e \uparrow\uparrow k
</math>
using [[Tetration|tetration]] or [[Knuth's up-arrow notation]].

To see the divergence of the series ({{EquationNote|4}}) using the integral test, note that by repeated application of the [[Chain rule|chain rule]]
:<math>
\frac{d}{dx}\ln_{k+1}(x)
=\frac{d}{dx}\ln(\ln_k(x))
=\frac1{\ln_k(x)}\frac{d}{dx}\ln_k(x)
=\cdots
=\frac1{x\ln(x)\cdots\ln_k(x)},
</math>
hence
:<math>
\int_{N_k}^\infty\frac{dx}{x\ln(x)\cdots\ln_k(x)}
=\ln_{k+1}(x)\bigr|_{N_k}^\infty=\infty.
</math>
To see the convergence of the series ({{EquationNote|5}}), note that by the [[Power rule|power rule]], the chain rule and the above result
:<math>
-\frac{d}{dx}\frac1{\varepsilon(\ln_k(x))^\varepsilon}
=\frac1{(\ln_k(x))^{1+\varepsilon}}\frac{d}{dx}\ln_k(x)
=\cdots
=\frac{1}{x\ln(x)\cdots\ln_{k-1}(x)(\ln_k(x))^{1+\varepsilon}},
</math>
hence
:<math>
\int_{N_k}^\infty\frac{dx}{x\ln(x)\cdots\ln_{k-1}(x)(\ln_k(x))^{1+\varepsilon}}
=-\frac1{\varepsilon(\ln_k(x))^\varepsilon}\biggr|_{N_k}^\infty<\infty
</math>
and ({{EquationNote|1}}) gives bounds for the infinite series in ({{EquationNote|5}}).

==See also==
*[[Convergence tests]]
*Convergence (mathematics)
*[[Direct comparison test]]
*[[Dominated convergence theorem]]
*Euler-Maclaurin formula
*[[Limit comparison test]]
*[[Monotone convergence theorem]]

==References==
* [[Biography:Konrad Knopp|Knopp, Konrad]], "Infinite Sequences and Series", [[Social:Dover Publications|Dover Publications]], Inc.,  New York, 1956. (§ 3.3) {{ISBN|0-486-60153-6}}
* [[Whittaker and Watson|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}}
<references/>

{{Calculus topics}}

[[Category:Integral calculus]]
[[Category:Convergence tests]]
[[Category:Articles containing proofs]]

{{Sourceattribution|Integral test for convergence}}