Английская Википедия:Integral test for convergence
In mathematics, the integral test for convergence is a method used to test infinite series of monotonous 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 Шаблон:Math and a function Шаблон:Math defined on the unbounded interval Шаблон:Closed-open, on which it is monotone decreasing. Then the infinite series
- <math>\sum_{n=N}^\infty f(n)</math>
converges to a real number if and only if the improper integral
- <math>\int_N^\infty f(x)\,dx</math>
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 <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 comparison test, comparing the term Шаблон:Math with the integral of Шаблон:Math over the intervals Шаблон:Closed-open and Шаблон:Closed-open, respectively.
The monotonous function <math>f</math> is continuous 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 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 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 that has the least index in an enumeration <math>\mathbb N\to\mathbb Q</math> and satisfies the above property. Since <math>f</math> is monotone, this defines an injective mapping <math>c:D\to\mathbb Q, x\mapsto c(x)</math> and thus <math>D</math> is countable. It follows that <math>f</math> is continuous almost everywhere. This is sufficient for Riemann integrability.[1]
Since Шаблон:Math 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 Шаблон:Math,
and, for every integer Шаблон:Math,
By summation over all Шаблон:Math from Шаблон:Math to some larger integer Шаблон:Math, we get from (Шаблон:EquationNote)
- <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>
and from (Шаблон:EquationNote)
- <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 Шаблон:Math tend to infinity, the bounds in (Шаблон:EquationNote) and the result follow.
Applications
The harmonic series
- <math>
\sum_{n=1}^\infty \frac 1 n </math> diverges because, using the natural logarithm, its antiderivative, and the 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 Шаблон:Math, because by the 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 (Шаблон:EquationNote) 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 Шаблон:Math decreases to 0 faster than Шаблон:Math but slower than Шаблон:Math 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 Шаблон:Math, and whether the corresponding series of the Шаблон:Math still diverges. Once such a sequence is found, a similar question can be asked with Шаблон:Math taking the role of Шаблон: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 Шаблон:Math, the series Шаблон:NumBlk still diverges (cf. proof that the sum of the reciprocals of the primes diverges for Шаблон:Math) but Шаблон:NumBlk </math>|Шаблон:EquationRef}} converges for every Шаблон:Math. Here Шаблон:Math denotes the Шаблон:Math-fold composition of the natural logarithm defined 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, Шаблон:Math denotes the smallest natural number such that the Шаблон:Math-fold composition is well-defined and Шаблон:Math, i.e.
- <math>
N_k\ge \underbrace{e^{e^{\cdot^{\cdot^{e}}}}}_{k\ e'\text{s}}=e \uparrow\uparrow k </math> using tetration or Knuth's up-arrow notation.
To see the divergence of the series (Шаблон:EquationNote) using the integral test, note that by repeated application of the 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 (Шаблон:EquationNote), note that by the 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 (Шаблон:EquationNote) gives bounds for the infinite series in (Шаблон: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) Шаблон:ISBN
- Whittaker, E. T., and Watson, G. N., A Course in Modern Analysis, fourth edition, Cambridge University Press, 1963. (§ 4.43) Шаблон:ISBN
- Ferreira, Jaime Campos, Ed Calouste Gulbenkian, 1987, Шаблон:ISBN
